Source

fcla / src / section-LISS.xml

Full commit
   1
   2
   3
   4
   5
   6
   7
   8
   9
  10
  11
  12
  13
  14
  15
  16
  17
  18
  19
  20
  21
  22
  23
  24
  25
  26
  27
  28
  29
  30
  31
  32
  33
  34
  35
  36
  37
  38
  39
  40
  41
  42
  43
  44
  45
  46
  47
  48
  49
  50
  51
  52
  53
  54
  55
  56
  57
  58
  59
  60
  61
  62
  63
  64
  65
  66
  67
  68
  69
  70
  71
  72
  73
  74
  75
  76
  77
  78
  79
  80
  81
  82
  83
  84
  85
  86
  87
  88
  89
  90
  91
  92
  93
  94
  95
  96
  97
  98
  99
 100
 101
 102
 103
 104
 105
 106
 107
 108
 109
 110
 111
 112
 113
 114
 115
 116
 117
 118
 119
 120
 121
 122
 123
 124
 125
 126
 127
 128
 129
 130
 131
 132
 133
 134
 135
 136
 137
 138
 139
 140
 141
 142
 143
 144
 145
 146
 147
 148
 149
 150
 151
 152
 153
 154
 155
 156
 157
 158
 159
 160
 161
 162
 163
 164
 165
 166
 167
 168
 169
 170
 171
 172
 173
 174
 175
 176
 177
 178
 179
 180
 181
 182
 183
 184
 185
 186
 187
 188
 189
 190
 191
 192
 193
 194
 195
 196
 197
 198
 199
 200
 201
 202
 203
 204
 205
 206
 207
 208
 209
 210
 211
 212
 213
 214
 215
 216
 217
 218
 219
 220
 221
 222
 223
 224
 225
 226
 227
 228
 229
 230
 231
 232
 233
 234
 235
 236
 237
 238
 239
 240
 241
 242
 243
 244
 245
 246
 247
 248
 249
 250
 251
 252
 253
 254
 255
 256
 257
 258
 259
 260
 261
 262
 263
 264
 265
 266
 267
 268
 269
 270
 271
 272
 273
 274
 275
 276
 277
 278
 279
 280
 281
 282
 283
 284
 285
 286
 287
 288
 289
 290
 291
 292
 293
 294
 295
 296
 297
 298
 299
 300
 301
 302
 303
 304
 305
 306
 307
 308
 309
 310
 311
 312
 313
 314
 315
 316
 317
 318
 319
 320
 321
 322
 323
 324
 325
 326
 327
 328
 329
 330
 331
 332
 333
 334
 335
 336
 337
 338
 339
 340
 341
 342
 343
 344
 345
 346
 347
 348
 349
 350
 351
 352
 353
 354
 355
 356
 357
 358
 359
 360
 361
 362
 363
 364
 365
 366
 367
 368
 369
 370
 371
 372
 373
 374
 375
 376
 377
 378
 379
 380
 381
 382
 383
 384
 385
 386
 387
 388
 389
 390
 391
 392
 393
 394
 395
 396
 397
 398
 399
 400
 401
 402
 403
 404
 405
 406
 407
 408
 409
 410
 411
 412
 413
 414
 415
 416
 417
 418
 419
 420
 421
 422
 423
 424
 425
 426
 427
 428
 429
 430
 431
 432
 433
 434
 435
 436
 437
 438
 439
 440
 441
 442
 443
 444
 445
 446
 447
 448
 449
 450
 451
 452
 453
 454
 455
 456
 457
 458
 459
 460
 461
 462
 463
 464
 465
 466
 467
 468
 469
 470
 471
 472
 473
 474
 475
 476
 477
 478
 479
 480
 481
 482
 483
 484
 485
 486
 487
 488
 489
 490
 491
 492
 493
 494
 495
 496
 497
 498
 499
 500
 501
 502
 503
 504
 505
 506
 507
 508
 509
 510
 511
 512
 513
 514
 515
 516
 517
 518
 519
 520
 521
 522
 523
 524
 525
 526
 527
 528
 529
 530
 531
 532
 533
 534
 535
 536
 537
 538
 539
 540
 541
 542
 543
 544
 545
 546
 547
 548
 549
 550
 551
 552
 553
 554
 555
 556
 557
 558
 559
 560
 561
 562
 563
 564
 565
 566
 567
 568
 569
 570
 571
 572
 573
 574
 575
 576
 577
 578
 579
 580
 581
 582
 583
 584
 585
 586
 587
 588
 589
 590
 591
 592
 593
 594
 595
 596
 597
 598
 599
 600
 601
 602
 603
 604
 605
 606
 607
 608
 609
 610
 611
 612
 613
 614
 615
 616
 617
 618
 619
 620
 621
 622
 623
 624
 625
 626
 627
 628
 629
 630
 631
 632
 633
 634
 635
 636
 637
 638
 639
 640
 641
 642
 643
 644
 645
 646
 647
 648
 649
 650
 651
 652
 653
 654
 655
 656
 657
 658
 659
 660
 661
 662
 663
 664
 665
 666
 667
 668
 669
 670
 671
 672
 673
 674
 675
 676
 677
 678
 679
 680
 681
 682
 683
 684
 685
 686
 687
 688
 689
 690
 691
 692
 693
 694
 695
 696
 697
 698
 699
 700
 701
 702
 703
 704
 705
 706
 707
 708
 709
 710
 711
 712
 713
 714
 715
 716
 717
 718
 719
 720
 721
 722
 723
 724
 725
 726
 727
 728
 729
 730
 731
 732
 733
 734
 735
 736
 737
 738
 739
 740
 741
 742
 743
 744
 745
 746
 747
 748
 749
 750
 751
 752
 753
 754
 755
 756
 757
 758
 759
 760
 761
 762
 763
 764
 765
 766
 767
 768
 769
 770
 771
 772
 773
 774
 775
 776
 777
 778
 779
 780
 781
 782
 783
 784
 785
 786
 787
 788
 789
 790
 791
 792
 793
 794
 795
 796
 797
 798
 799
 800
 801
 802
 803
 804
 805
 806
 807
 808
 809
 810
 811
 812
 813
 814
 815
 816
 817
 818
 819
 820
 821
 822
 823
 824
 825
 826
 827
 828
 829
 830
 831
 832
 833
 834
 835
 836
 837
 838
 839
 840
 841
 842
 843
 844
 845
 846
 847
 848
 849
 850
 851
 852
 853
 854
 855
 856
 857
 858
 859
 860
 861
 862
 863
 864
 865
 866
 867
 868
 869
 870
 871
 872
 873
 874
 875
 876
 877
 878
 879
 880
 881
 882
 883
 884
 885
 886
 887
 888
 889
 890
 891
 892
 893
 894
 895
 896
 897
 898
 899
 900
 901
 902
 903
 904
 905
 906
 907
 908
 909
 910
 911
 912
 913
 914
 915
 916
 917
 918
 919
 920
 921
 922
 923
 924
 925
 926
 927
 928
 929
 930
 931
 932
 933
 934
 935
 936
 937
 938
 939
 940
 941
 942
 943
 944
 945
 946
 947
 948
 949
 950
 951
 952
 953
 954
 955
 956
 957
 958
 959
 960
 961
 962
 963
 964
 965
 966
 967
 968
 969
 970
 971
 972
 973
 974
 975
 976
 977
 978
 979
 980
 981
 982
 983
 984
 985
 986
 987
 988
 989
 990
 991
 992
 993
 994
 995
 996
 997
 998
 999
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009
1010
1011
1012
1013
1014
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
1061
1062
1063
1064
1065
1066
1067
1068
1069
1070
1071
1072
1073
1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
1105
1106
1107
1108
1109
1110
1111
1112
1113
1114
1115
1116
1117
1118
1119
1120
1121
1122
1123
1124
1125
1126
1127
1128
1129
1130
1131
1132
1133
1134
1135
1136
1137
1138
1139
1140
1141
1142
1143
1144
1145
1146
1147
1148
1149
1150
1151
1152
1153
1154
1155
1156
1157
1158
<?xml version="1.0" encoding="UTF-8" ?>
<section acro="LISS">
<title>Linear Independence and Spanning Sets</title>

<!-- %%%%%%%%%% -->
<!-- % -->
<!-- %  Section LISS -->
<!-- %  Linear Independence and Spanning Sets -->
<!-- % -->
<!-- %%%%%%%%%% -->
<introduction>
<p>A vector space is defined as a set with two operations, meeting ten properties (<acroref type="definition" acro="VS" />).  Just as the definition of span of a set of vectors only required knowing how to add vectors and how to multiply vectors by scalars, so it is with linear independence.  A definition of a linear independent set of vectors in an arbitrary vector space only requires knowing how to form linear combinations and equating these with the zero vector.  Since every vector space must have a zero vector (<acroref type="property" acro="Z" />), we always have a zero vector at our disposal.</p>

<p>In this section we will also put a twist on the notion of the span of a set of vectors.  Rather than beginning with a set of vectors and creating a subspace that is the span, we will instead begin with a subspace and look for a set of vectors whose span equals the subspace.</p>

<p>The combination of linear independence and spanning will be very important going forward.
</p>

</introduction>

<subsection acro="LI">
<title>Linear Independence</title>

<p>Our previous definition of linear independence (<acroref type="definition" acro="LICV" />) employed a relation of linear dependence that was a linear combination on one side of an equality and a zero vector on the other side.  As a linear combination in a vector space (<acroref type="definition" acro="LC" />) depends only on vector addition and scalar multiplication, and every vector space must have a zero vector (<acroref type="property" acro="Z" />), we can extend our definition of linear independence from the setting of $\complex{m}$ to the setting of a general vector space $V$ with almost no changes.  Compare these next two definitions with <acroref type="definition" acro="RLDCV" /> and <acroref type="definition" acro="LICV" />.</p>

<definition acro="RLD" index="relation of linear dependence">
<title>Relation of Linear Dependence</title>
<p>Suppose that $V$ is a vector space.
Given a set of vectors $S=\set{\vectorlist{u}{n}}$, an equation of the form
<equation>
\lincombo{\alpha}{u}{n}=\zerovector
</equation>
is a <define>relation of linear dependence</define> on $S$.  If this equation is formed in a trivial fashion, <ie /> $\alpha_i=0$, $1\leq i\leq n$, then we say it is a <define>trivial relation of linear dependence</define> on $S$.
</p>

</definition>

<definition acro="LI" index="linear independence">
<title>Linear Independence</title>
<p>Suppose that $V$ is a vector space.
The set of vectors $S=\set{\vectorlist{u}{n}}$ from $V$ is <define>linearly dependent</define> if there is a relation of linear dependence on $S$ that is not trivial.  In the case where the <em>only</em> relation of linear dependence on $S$ is the trivial one, then $S$ is a <define>linearly independent</define> set of vectors.</p>

</definition>

<p>Notice the emphasis on the word <q>only.</q>  This might remind you of the definition of a nonsingular matrix, where if the matrix is employed as the coefficient matrix of a homogeneous system then the <em>only</em> solution is the <em>trivial</em> one.</p>

<example acro="LIP4" index="linearly independent!polynomials">
<title>Linear independence in $P_4$</title>

<p>In the vector space of polynomials with degree 4 or less, $P_4$ (<acroref type="example" acro="VSP" />) consider the set $S$ below
<alignmath>
\set{
2x^4+3x^3+2x^2-x+10,\,
-x^4-2x^3+x^2+5x-8,\,
2x^4+x^3+10x^2+17x-2
}
</alignmath>
</p>

<p>Is this set of vectors linearly independent or dependent?  Consider that
<alignmath>
<![CDATA[&3\left(2x^4+3x^3+2x^2-x+10\right)]]>
+4\left(-x^4-2x^3+x^2+5x-8\right)\\
<![CDATA[&\quad +(-1)\left(2x^4+x^3+10x^2+17x-2\right)]]>
=0x^4+0x^3+0x^2+0x+0=\zerovector
</alignmath>
This is a nontrivial relation of linear dependence (<acroref type="definition" acro="RLD" />) on the set $S$ and so convinces us that $S$ is linearly dependent (<acroref type="definition" acro="LI" />).</p>

<p>Now, I hear you say, <q>Where did <em>those</em> scalars come from?</q>  Do not worry about that right now, just be sure you understand why the above explanation is sufficient to prove that $S$ is linearly dependent.  The remainder of the example will demonstrate how we might find these scalars if they had not been provided so readily.</p>

<p>Let's look at another set of vectors (polynomials) from $P_4$.  Let
<alignmath>
<![CDATA[T&=\left\{]]>
3x^4-2x^3+4x^2+6x-1,\,
-3x^4+1x^3+0x^2+4x+2,\right.\\
<![CDATA[&\quad \left.4x^4+5x^3-2x^2+3x+1,\,]]>
2x^4-7x^3+4x^2+2x+1\right\}
</alignmath>
</p>

<p>Suppose we have a relation of linear dependence on this set,
<alignmath>
<![CDATA[\zerovector&=0x^4+0x^3+0x^2+0x+0\\]]>
<![CDATA[&=\alpha_1\left(3x^4-2x^3+4x^2+6x-1\right)+\alpha_2\left(-3x^4+1x^3+0x^2+4x+2\right)\\]]>
<![CDATA[&\quad +\alpha_3\left(4x^4+5x^3-2x^2+3x+1\right)+\alpha_4\left(2x^4-7x^3+4x^2+2x+1\right)]]>
</alignmath>
</p>

<p>Using our definitions of vector addition and scalar multiplication in $P_4$ (<acroref type="example" acro="VSP" />), we arrive at,
<alignmath>
<![CDATA[&0x^4+0x^3+0x^2+0x+0=\\]]>
<![CDATA[&\quad\left(3\alpha_1-3\alpha_2+4\alpha_3+2\alpha_4\right)x^4 + \left(-2\alpha_1+\alpha_2+5\alpha_3-7\alpha_4\right)x^3 +\ \\]]>
<![CDATA[&\quad\left(4\alpha_1+              -2\alpha_3+4\alpha_4\right)x^2+\left(6\alpha_1+4\alpha_2+3\alpha_3+2\alpha_4\right)x + \left(-\alpha_1+2\alpha_2+\alpha_3+\alpha_4\right)]]>
</alignmath>
</p>

<p>Equating coefficients, we arrive at the homogeneous system of equations,
<alignmath>
<![CDATA[3\alpha_1-3\alpha_2+4\alpha_3+2\alpha_4&=0\\]]>
<![CDATA[-2\alpha_1+\alpha_2+5\alpha_3-7\alpha_4&=0\\]]>
<![CDATA[4\alpha_1+              -2\alpha_3+4\alpha_4&=0\\]]>
<![CDATA[6\alpha_1+4\alpha_2+3\alpha_3+2\alpha_4&=0\\]]>
<![CDATA[-\alpha_1+2\alpha_2+\alpha_3+\alpha_4&=0]]>
</alignmath>
</p>

<p>We form the coefficient matrix of this homogeneous system of equations and row-reduce to find
<equation>
\begin{bmatrix}
<![CDATA[\leading{1} & 0 & 0 & 0\\]]>
<![CDATA[0 & \leading{1} & 0 & 0\\]]>
<![CDATA[0 & 0 & \leading{1} & 0\\]]>
<![CDATA[0 & 0 & 0 & \leading{1}\\]]>
<![CDATA[0 & 0 & 0 & 0]]>
\end{bmatrix}
</equation>
</p>

<p>We expected the system to be consistent (<acroref type="theorem" acro="HSC" />) and so can compute $n-r=4-4=0$ and <acroref type="theorem" acro="CSRN" /> tells us that the solution is unique.  Since this is a homogeneous system, this unique solution is the trivial solution (<acroref type="definition" acro="TSHSE" />),  $\alpha_1=0$, $\alpha_2=0$, $\alpha_3=0$, $\alpha_4=0$.  So by <acroref type="definition" acro="LI" /> the set $T$ is linearly independent.</p>

<p>A few observations.  If we had discovered infinitely many solutions, then we could have used one of the non-trivial solutions to provide a linear combination in the manner we used to show that $S$ was linearly dependent.  It is important to realize that it is not interesting that we can create a relation of linear dependence with zero scalars <mdash /> we can <em>always</em> do that <mdash /> but for $T$, this is the <em>only</em> way to create a relation of linear dependence.  It was no accident that we arrived at a homogeneous system of equations in this example, it is related to our use of the zero vector in defining a relation of linear dependence.  It is easy to present a convincing statement that a set is linearly dependent (just exhibit a nontrivial relation of linear dependence) but a convincing statement of linear independence requires demonstrating that there is no relation of linear dependence other than the trivial one.  Notice how we relied on theorems from <acroref type="chapter" acro="SLE" /> to provide this demonstration.  Whew!  There's a lot going on in this example.  Spend some time with it, we'll be waiting patiently right here when you get back.</p>

</example>

<example acro="LIM32" index="linear independence!matrices">
<title>Linear independence in $M_{32}$</title>

<p>Consider the two sets of vectors $R$ and $S$ from the vector space of all $3\times 2$ matrices, $M_{32}$ (<acroref type="example" acro="VSM" />)
<alignmath>
<![CDATA[R&=\set{]]>
\begin{bmatrix}
<![CDATA[3 & -1\\1 & 4\\6 & -6]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[-2 & 3\\1 & -3\\-2 & -6]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[6 & -6\\-1 & 0\\7 & -9]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[7 & 9\\-4 & -5\\2 & 5]]>
\end{bmatrix}
}\\
<![CDATA[S&=\set{]]>
\begin{bmatrix}
<![CDATA[2 & 0\\ 1 & -1\\ 1 & 3]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[-4 & 0\\ -2 & 2\\ -2 & -6]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[1 & 1\\ -2 & 1\\ 2 & 4]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[-5 & 3\\ -10 & 7\\ 2 & 0]]>
\end{bmatrix}
}
</alignmath></p>

<p>One set is linearly independent, the other is not.  Which is which?  Let's examine $R$ first.  Build a generic relation of linear dependence (<acroref type="definition" acro="RLD" />),
<equation>
\alpha_1\begin{bmatrix}
<![CDATA[3 & -1\\1 & 4\\6 & -6]]>
\end{bmatrix}+
\alpha_2\begin{bmatrix}
<![CDATA[-2 & 3\\1 & -3\\-2 & -6]]>
\end{bmatrix}+
\alpha_3\begin{bmatrix}
<![CDATA[6 & -6\\-1 & 0\\7 & -9]]>
\end{bmatrix}+
\alpha_4\begin{bmatrix}
<![CDATA[7 & 9\\-4 & -5\\2 & 5]]>
\end{bmatrix}=
\zerovector
</equation>
</p>

<p>Massaging the left-hand side with our definitions of vector addition and scalar multiplication in $M_{32}$ (<acroref type="example" acro="VSM" />) we obtain,
<equation>
\begin{bmatrix}
<![CDATA[3\alpha_1-2\alpha_2+6\alpha_3+7\alpha_4 &]]>
-\alpha_1+3\alpha_2-6\alpha_3+9\alpha_4 \\
<![CDATA[\alpha_1+\alpha_2-\alpha_3-4\alpha_4 &]]>
4\alpha_1-3\alpha_2+            -5\alpha_4 \\
<![CDATA[6\alpha_1-2\alpha_2+7\alpha_3+2\alpha_4 &]]>
-6\alpha_1-6\alpha_2-9\alpha_3+5\alpha_4
\end{bmatrix}
=\begin{bmatrix}
<![CDATA[0&0\\0&0\\0&0]]>
\end{bmatrix}
</equation>
</p>

<p>Using our definition of matrix equality (<acroref type="definition" acro="ME" />) and equating corresponding entries we get the homogeneous system of six equations in four variables,
<alignmath>
<![CDATA[3\alpha_1-2\alpha_2+6\alpha_3+7\alpha_4&=0\\]]>
<![CDATA[-\alpha_1+3\alpha_2-6\alpha_3+9\alpha_4&=0\\]]>
<![CDATA[\alpha_1+\alpha_2-\alpha_3-4\alpha_4&=0\\]]>
<![CDATA[4\alpha_1-3\alpha_2+            -5\alpha_4&=0\\]]>
<![CDATA[6\alpha_1-2\alpha_2+7\alpha_3+2\alpha_4&=0\\]]>
<![CDATA[-6\alpha_1-6\alpha_2-9\alpha_3+5\alpha_4&=0]]>
</alignmath>
</p>

<p>Form the coefficient matrix of this homogeneous system and row-reduce to obtain
<equation>
\begin{bmatrix}
<![CDATA[\leading{1} & 0 & 0 & 0\\]]>
<![CDATA[0 & \leading{1} & 0 & 0\\]]>
<![CDATA[0 & 0 & \leading{1} & 0\\]]>
<![CDATA[0 & 0 & 0 & \leading{1}\\]]>
<![CDATA[0 & 0 & 0 & 0\\]]>
<![CDATA[0 & 0 & 0 & 0]]>
\end{bmatrix}
</equation>
</p>

<p>Analyzing this matrix we are led to conclude that $\alpha_1=0$, $\alpha_2=0$, $\alpha_3=0$, $\alpha_4=0$.  This means there is <em>only</em> a trivial relation of linear dependence on the vectors of $R$ and so we call $R$ a linearly independent set (<acroref type="definition" acro="LI" />).</p>

<p>So it must be that $S$ is linearly dependent.  Let's see if we can find a non-trivial relation of linear dependence on $S$.  We will begin as with $R$, by constructing a relation of linear dependence (<acroref type="definition" acro="RLD" />) with unknown scalars,
<equation>
\alpha_1\begin{bmatrix}
<![CDATA[2 & 0\\ 1 & -1\\ 1 & 3]]>
\end{bmatrix}+
\alpha_2\begin{bmatrix}
<![CDATA[-4 & 0\\ -2 & 2\\ -2 & -6]]>
\end{bmatrix}+
\alpha_3\begin{bmatrix}
<![CDATA[1 & 1\\ -2 & 1\\ 2 & 4]]>
\end{bmatrix}+
\alpha_4\begin{bmatrix}
<![CDATA[-5 & 3\\ -10 & 7\\ 2 & 0]]>
\end{bmatrix}=
\zerovector
</equation>
</p>

<p>Massaging the left-hand side with our definitions of vector addition and scalar multiplication in $M_{32}$ (<acroref type="example" acro="VSM" />) we obtain,
<equation>
\begin{bmatrix}
<![CDATA[2\alpha_1-4\alpha_2+\alpha_3-5\alpha_4&]]>
                              \alpha_3+3\alpha_4\\
<![CDATA[\alpha_1-2\alpha_2-2\alpha_3-10\alpha_4&]]>
-\alpha_1+2\alpha_2+\alpha_3+7\alpha_4\\
<![CDATA[\alpha_1-2\alpha_2+2\alpha_3+2\alpha_4&]]>
3\alpha_1-6\alpha_2+4\alpha_3
\end{bmatrix}
=\begin{bmatrix}
<![CDATA[0&0\\0&0\\0&0]]>
\end{bmatrix}
</equation>
</p>

<p>Using our definition of matrix equality (<acroref type="definition" acro="ME" />) and equating corresponding entries we get the homogeneous system of six equations in four variables,
<alignmath>
<![CDATA[2\alpha_1-4\alpha_2+\alpha_3-5\alpha_4&=0\\]]>
<![CDATA[                             +\alpha_3+3\alpha_4&=0\\]]>
<![CDATA[\alpha_1-2\alpha_2-2\alpha_3-10\alpha_4&=0\\]]>
<![CDATA[-\alpha_1+2\alpha_2+\alpha_3+7\alpha_4&=0\\]]>
<![CDATA[\alpha_1-2\alpha_2+2\alpha_3+2\alpha_4&=0\\]]>
<![CDATA[3\alpha_1-6\alpha_2+4\alpha_3         &=0]]>
</alignmath>
</p>

<p>Form the coefficient matrix of this homogeneous system and row-reduce to obtain
<equation>
\begin{bmatrix}
<![CDATA[\leading{1} & -2 & 0 & -4\\]]>
<![CDATA[0 & 0 & \leading{1} & 3\\]]>
<![CDATA[0 & 0 & 0 & 0\\]]>
<![CDATA[0 & 0 & 0 & 0\\]]>
<![CDATA[0 & 0 & 0 & 0\\]]>
<![CDATA[0 & 0 & 0 & 0]]>
\end{bmatrix}
</equation>
</p>

<p>Analyzing this we see that the system is consistent (we expected this since the system is homogeneous, <acroref type="theorem" acro="HSC" />) and has $n-r=4-2=2$ free variables, namely $\alpha_2$ and $\alpha_4$.  This means there are infinitely many solutions, and in particular, we can find a non-trivial solution, so long as we do not pick all of our free variables to be zero.  The mere presence of a nontrivial solution for these scalars is enough to conclude that  $S$ is a linearly dependent set (<acroref type="definition" acro="LI" />).  But let's go ahead and explicitly construct a non-trivial relation of linear dependence.</p>

<p>Choose $\alpha_2=1$ and $\alpha_4=-1$.  There is nothing special about this choice, there are infinitely many possibilities, some <q>easier</q> than this one, just avoid picking both variables to be zero.  Then we find the corresponding dependent variables to be $\alpha_1=-2$ and $\alpha_3=3$.  So the relation of linear dependence,
<equation>
(-2)\begin{bmatrix}
<![CDATA[2 & 0\\ 1 & -1\\ 1 & 3]]>
\end{bmatrix}+
(1)\begin{bmatrix}
<![CDATA[-4 & 0\\ -2 & 2\\ -2 & -6]]>
\end{bmatrix}+
(3)\begin{bmatrix}
<![CDATA[1 & 1\\ -2 & 1\\ 2 & 4]]>
\end{bmatrix}+
(-1)\begin{bmatrix}
<![CDATA[-5 & 3\\ -10 & 7\\ 2 & 0]]>
\end{bmatrix}
=
\begin{bmatrix}
<![CDATA[0&0\\0&0\\0&0]]>
\end{bmatrix}
</equation>
is an iron-clad demonstration that $S$ is linearly dependent.  Can you construct another such demonstration?
</p>

</example>

<example acro="LIC" index="linearly independent!crazy vector space">
<title>Linearly independent set in the crazy vector space</title>

<p>Is the set $R=\set{(1,\,0),\,(6,\,3)}$ linearly independent in the crazy vector space $C$ (<acroref type="example" acro="CVS" />)?</p>

<p>We begin with an arbitrary relation of linear dependence on $R$
<alignmath>
<![CDATA[\zerovector &= a_1(1,\,0) + a_2(6,\,3)&&]]>\text{<acroref type="definition" acro="RLD" />}\\
</alignmath>
and then massage it to a point where we can apply the definition of equality in $C$.  Recall the definitions of vector addition and scalar multiplication in $C$ are not what you would expect.
<alignmath>
<![CDATA[(&-1,\,-1)\\]]>
<![CDATA[&=\zerovector]]>
<![CDATA[&&]]>\text{<acroref type="example" acro="CVS" />}\\
<![CDATA[&=a_1(1,\,0) + a_2(6,\,3)]]>
<![CDATA[&&]]>\text{<acroref type="definition" acro="RLD" />}\\
<![CDATA[&=(1a_1+a_1-1,\,0a_1+a_1-1) + (6a_2+a_2-1,\,3a_2+a_2-1)]]>
<![CDATA[&&]]>\text{<acroref type="example" acro="CVS" />}\\
<![CDATA[&=(2a_1-1,\,a_1-1) + (7a_2-1,\,4a_2-1)\\]]>
<![CDATA[&=(2a_1-1+7a_2-1+1,\,a_1-1+4a_2-1+1)]]>
<![CDATA[&&]]>\text{<acroref type="example" acro="CVS" />}\\
<![CDATA[&=(2a_1+7a_2-1,\,a_1+4a_2-1)]]>
</alignmath>
</p>

<p>Equality in $C$ (<acroref type="example" acro="CVS" />) then yields the two equations,
<alignmath>
<![CDATA[2a_1+7a_2-1&=-1\\]]>
<![CDATA[a_1+4a_2-1&=-1]]>
</alignmath>
which becomes the homogeneous system
<alignmath>
<![CDATA[2a_1+7a_2&=0\\]]>
<![CDATA[a_1+4a_2&=0]]>
</alignmath>
</p>

<p>Since the coefficient matrix of this system is nonsingular (check this!) the system has only the trivial solution $a_1=a_2=0$.  By <acroref type="definition" acro="LI" /> the set $R$ is linearly independent.  Notice that even though the zero vector of $C$ is not what we might first suspected, a question about linear independence still concludes with a question about a homogeneous system of equations.  Hmmm.</p>

</example>

</subsection>

<subsection acro="SS">
<title>Spanning Sets</title>

<p>In a vector space $V$, suppose we are given a set of vectors $S\subseteq V$.  Then we can immediately construct a subspace, $\spn{S}$, using <acroref type="definition" acro="SS" /> and then be assured by <acroref type="theorem" acro="SSS" /> that the construction does provide a subspace.  We now turn the situation upside-down.  Suppose we are first given a subspace $W\subseteq V$.  Can we find a set $S$ so that $\spn{S}=W$?  Typically $W$ is infinite and we are searching for a finite set of vectors $S$ that we can combine in linear combinations and <q>build</q> all of $W$.</p>

<p>I like to think of $S$ as the raw materials that are sufficient for the construction of $W$.  If you have nails, lumber, wire, copper pipe, drywall, plywood, carpet, shingles, paint (and a few other things), then you can combine them in many different ways to create a house (or infinitely many different houses for that matter).  A fast-food restaurant may have beef, chicken, beans, cheese, tortillas, taco shells and hot sauce and from this small list of ingredients build a wide variety of items for sale.  Or maybe a better analogy comes from Ben Cordes <mdash /> the additive primary colors (red, green and blue) can be combined to create many different colors by varying the intensity of each.  The intensity is like a scalar multiple, and the combination of the three intensities is like vector addition.  The three individual colors, red, green and blue, are the elements of the spanning set.</p>

<p>Because we will use terms like <q>spanned by</q> and <q>spanning set,</q> there is the potential for confusion with <q>the span.</q>  Come back and reread the first paragraph of this subsection whenever you are uncertain about the difference.  Here's the working definition.</p>

<definition acro="SSVS" index="spanning set">
<title>Spanning Set of a Vector Space</title>
<p>Suppose $V$ is a vector space.  A subset $S$ of $V$ is a <define>spanning set</define> of $V$ if $\spn{S}=V$.  In this case, we also frequently say $S$ <define>spans</define> $V$.</p>

</definition>

<p>The definition of a spanning set requires that two sets (subspaces actually) be equal.  If $S$ is a subset of $V$, then $\spn{S}\subseteq V$, always.  Thus it is usually only necessary to prove that $V\subseteq\spn{S}$.  Now would be a good time to review <acroref type="definition" acro="SE" />.</p>

<example acro="SSP4" index="spanning set!polynomials">
<title>Spanning set in $P_4$</title>

<p>In <acroref type="example" acro="SP4" /> we showed that
<equation>
W=\setparts{p(x)}{p\in P_4,\ p(2)=0}
</equation>
is a subspace of $P_4$, the vector space of polynomials with degree at most $4$ (<acroref type="example" acro="VSP" />).  In this example, we will show that the set
<equation>
S=\set{x-2,\,x^2-4x+4,\,x^3-6x^2+12x-8,\,x^4-8x^3+24x^2-32x+16}
</equation>
is a spanning set for $W$.  To do this, we require that $W=\spn{S}$.  This is an equality of sets.  We can check that every polynomial in $S$ has $x=2$ as a root and therefore $S\subseteq W$.  Since $W$ is closed under addition and scalar multiplication, $\spn{S}\subseteq W$ also.</p>

<p>So it remains to show that $W\subseteq \spn{S}$ (<acroref type="definition" acro="SE" />).  To do this, begin by choosing an arbitrary polynomial in $W$, say $r(x)=ax^4+bx^3+cx^2+dx+e\in W$.  This polynomial is not as arbitrary as it would appear, since we also know it must have $x=2$ as a root.  This translates to
<equation>
0=a(2)^4+b(2)^3+c(2)^2+d(2)+e=16a+8b+4c+2d+e
</equation>
as a condition on $r$.</p>

<p>We wish to show that $r$ is a polynomial in $\spn{S}$, that is, we want to show that $r$ can be written as a linear combination of the vectors (polynomials) in $S$.  So let's try.
<alignmath>
<![CDATA[r(x)&=ax^4+bx^3+cx^2+dx+e\\]]>
<![CDATA[&=\alpha_1\left(x-2\right)+\alpha_2\left(x^2-4x+4\right)+\alpha_3\left(x^3-6x^2+12x-8\right)\\]]>
<![CDATA[&\quad +\alpha_4\left(x^4-8x^3+24x^2-32x+16\right)\\]]>
<![CDATA[&=\alpha_4x^4+]]>
\left(\alpha_3-8\alpha_4\right)x^3+
\left(\alpha_2-6\alpha_3+24\alpha_4\right)x^2\\
<![CDATA[&\quad +]]>
\left(\alpha_1-4\alpha_2+12\alpha_3-32\alpha_4\right)x+
\left(-2\alpha_1+4\alpha_2-8\alpha_3+16\alpha_4\right)
</alignmath>
</p>

<p>Equating coefficients (vector equality in $P_4$) gives the system of five equations in four variables,
<alignmath>
<![CDATA[\alpha_4&=a\\]]>
<![CDATA[\alpha_3-8\alpha_4&=b\\]]>
<![CDATA[\alpha_2-6\alpha_3+24\alpha_4&=c\\]]>
<![CDATA[\alpha_1-4\alpha_2+12\alpha_3-32\alpha_4&=d\\]]>
<![CDATA[-2\alpha_1+4\alpha_2-8\alpha_3+16\alpha_4&=e\\]]>
</alignmath>
</p>

<p>Any solution to this system of equations will provide the linear combination we need to determine if $r\in\spn{S}$, but we need to be convinced there is a solution for any values of $a,\,b,\,c,\,d,\,e$ that qualify $r$ to be a member of $W$.  So the question is:  is this system of equations consistent?  We will form the augmented matrix, and row-reduce. (We probably need to do this by hand, since the matrix is symbolic <mdash /> reversing the order of the first four rows is the best way to start).  We obtain a matrix in reduced row-echelon form
<alignmath>
<![CDATA[&\begin{bmatrix}]]>
<![CDATA[\leading{1}&0&0&0&32a+12b+4c+d\\]]>
<![CDATA[0&\leading{1}&0&0&24a+6b+c\\]]>
<![CDATA[0&0&\leading{1}&0&8a+b\\]]>
<![CDATA[0&0&0&\leading{1}&a\\]]>
<![CDATA[0&0&0&0&16a+8b+4c+2d+e]]>
\end{bmatrix}\\
<![CDATA[=&]]>
\begin{bmatrix}
<![CDATA[\leading{1}&0&0&0&32a+12b+4c+d\\]]>
<![CDATA[0&\leading{1}&0&0&24a+6b+c\\]]>
<![CDATA[0&0&\leading{1}&0&8a+b\\]]>
<![CDATA[0&0&0&\leading{1}&a\\]]>
<![CDATA[0&0&0&0&0]]>
\end{bmatrix}
</alignmath>
</p>

<p>For your results to match our first matrix, you may find it necessary to multiply the final row of your row-reduced matrix by the appropriate scalar, and/or add multiples of this row to some of the other rows.  To obtain the second version of the matrix, the last entry of the last column has been simplified to zero according to the one condition we were able to impose on an arbitrary polynomial from $W$.    So with no leading 1's in the last column, <acroref type="theorem" acro="RCLS" /> tells us this system is consistent.  Therefore, <em>any</em> polynomial from $W$ can be written as a linear combination of the polynomials in $S$, so $W\subseteq\spn{S}$. Therefore,  $W=\spn{S}$ and $S$ is a spanning set for $W$ by <acroref type="definition" acro="SSVS" />.</p>

<p>Notice that an alternative to row-reducing the augmented matrix by hand would be to appeal to <acroref type="theorem" acro="FS" /> by expressing the column space of the coefficient matrix as a null space, and then verifying that the condition on $r$ guarantees that $r$ is in the column space, thus implying that the system is always consistent.  Give it a try, we'll wait.  This has been a complicated example, but worth studying carefully.</p>

</example>

<p>Given a subspace and a set of vectors, as in <acroref type="example" acro="SSP4" /> it can take some work to determine that the set actually is a spanning set.  An even harder problem is to be confronted with a subspace and required to construct a spanning set with no guidance.  We will now work  an example of this flavor, but some of the steps will be unmotivated.  Fortunately, we will have some better tools for this type of problem later on.</p>

<example acro="SSM22" index="spanning set!matrices">
<title>Spanning set in $M_{22}$</title>

<p>In the space of all $2\times 2$ matrices, $M_{22}$ consider the subspace
<equation>
<![CDATA[Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0}]]>
</equation>
and find a spanning set for $Z$.</p>

<p>
<![CDATA[We need to construct a limited number of matrices in $Z$ so that every matrix in $Z$ can be expressed as a linear combination of this limited number of matrices.  Suppose that $B=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is a matrix in $Z$.  Then we can form a column vector with the entries of $B$ and write]]>
<equation>
\colvector{a\\b\\c\\d}\in
<![CDATA[\nsp{\begin{bmatrix}1 & 3 & -1 & -5\\-2 & -6 & 3 & 14\end{bmatrix}}]]>
</equation>
</p>

<p>Row-reducing this matrix and applying <acroref type="theorem" acro="REMES" /> we obtain the equivalent statement,
<equation>
\colvector{a\\b\\c\\d}\in
<![CDATA[\nsp{\begin{bmatrix}\leading{1} & 3 & 0 & -1\\0 & 0 & \leading{1} & 4\end{bmatrix}}]]>
</equation>
</p>

<p>We can then express the subspace $Z$ in the following equal forms,
<alignmath>
<![CDATA[Z&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0}\\]]>
<![CDATA[&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-d=0,\ c+4d=0}\\]]>
<![CDATA[&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a=-3b+d,\ c=-4d}\\]]>
<![CDATA[&=\setparts{\begin{bmatrix}-3b+d&b\\-4d&d\end{bmatrix}}{b,\,d\in\complex{\null}}\\]]>
<![CDATA[&=\setparts{]]>
<![CDATA[\begin{bmatrix}-3b&b\\0&0\end{bmatrix}+]]>
<![CDATA[\begin{bmatrix}d&0\\-4d&d\end{bmatrix}]]>
}{b,\,d\in\complex{\null}}\\
<![CDATA[&=\setparts{]]>
<![CDATA[b\begin{bmatrix}-3&1\\0&0\end{bmatrix}+]]>
<![CDATA[d\begin{bmatrix}1&0\\-4&1\end{bmatrix}]]>
}{b,\,d\in\complex{\null}}\\
<![CDATA[&=\spn{\set{]]>
<![CDATA[\begin{bmatrix}-3&1\\0&0\end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix}1&0\\-4&1\end{bmatrix}]]>
}}
</alignmath>
</p>

<p>So the set
<equation>
Q=\set{
<![CDATA[\begin{bmatrix}-3&1\\0&0\end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix}1&0\\-4&1\end{bmatrix}]]>
}
</equation>
spans $Z$ by <acroref type="definition" acro="SSVS" />.</p>

</example>

<example acro="SSC" index="spanning set!crazy vector space">
<title>Spanning set in the crazy vector space</title>

<p>In <acroref type="example" acro="LIC" /> we determined that the set $R=\set{(1,\,0),\,(6,\,3)}$ is linearly independent in the crazy vector space $C$ (<acroref type="example" acro="CVS" />).  We now show that $R$ is a spanning set for $C$.</p>

<p>Given an arbitrary vector $(x,\,y)\in C$ we desire to show that it can be written as a linear combination of the elements of $R$.  In other words, are there scalars $a_1$ and $a_2$ so that
<equation>
(x,\,y)=a_1(1,\,0) + a_2(6,\,3)
</equation>
</p>

<p>We will act as if this equation is true and try to determine just what $a_1$ and $a_2$ would be (as functions of $x$ and $y$).  Recall that our vector space operations are unconventional and are defined in <acroref type="example" acro="CVS" />.
<alignmath>
<![CDATA[(x,\,y)&=a_1(1,\,0) + a_2(6,\,3)\\]]>
<![CDATA[&= (1a_1+a_1-1,\,0a_1+a_1-1) + (6a_2+a_2-1,\,3a_2+a_2-1)\\]]>
<![CDATA[&= (2a_1-1,\,a_1-1) + (7a_2-1,\,4a_2-1)\\]]>
<![CDATA[&= (2a_1-1+7a_2-1+1,\,a_1-1+4a_2-1+1)\\]]>
<![CDATA[&= (2a_1+7a_2-1,\,a_1+4a_2-1)]]>
</alignmath>
</p>

<p>Equality in $C$ then yields the two equations,
<alignmath>
<![CDATA[2a_1+7a_2-1&=x\\]]>
<![CDATA[a_1+4a_2-1&=y]]>
</alignmath>
which becomes the linear system with a matrix representation
<equation>
\begin{bmatrix}
<![CDATA[2 & 7 \\ 1 & 4]]>
\end{bmatrix}
\colvector{a_1\\a_2}
=
\colvector{x+1\\y+1}
</equation>
</p>

<p>The coefficient matrix of this system is nonsingular, hence invertible (<acroref type="theorem" acro="NI" />), and we can employ its inverse to find a solution (<acroref type="theorem" acro="TTMI" />, <acroref type="theorem" acro="SNCM" />),
<equation>
\colvector{a_1\\a_2}=
<![CDATA[\inverse{\begin{bmatrix} 2 & 7 \\ 1 & 4 \end{bmatrix}}\colvector{x+1\\y+1}=]]>
<![CDATA[\begin{bmatrix} 4 & -7 \\ -1 & 2 \end{bmatrix}\colvector{x+1\\y+1}=]]>
\colvector{4x-7y-3\\-x+2y+1}
</equation>
</p>

<p>We could chase through the above implications backwards and take the existence of these solutions as sufficient evidence for $R$ being a spanning set for $C$.  Instead, let us view the above as simply scratchwork and now get serious with a simple direct proof that $R$ is a spanning set.  Ready?  Suppose $(x,\,y)$ is any vector from $C$, then compute the following linear combination using the definitions of the operations in $C$,
<alignmath>
<![CDATA[(4x&-7y-3)(1,\,0)+(-x+2y+1)(6,\,3)\\]]>
<![CDATA[&=\left(1(4x-7y-3)+(4x-7y-3)-1,\,0(4x-7y-3)+(4x-7y-3)-1\right)+\\]]>
<![CDATA[&\quad\left(6(-x+2y+1)+(-x+2y+1)-1,\,3(-x+2y+1)+(-x+2y+1)-1\right)\\]]>
<![CDATA[&=(8x-14y-7,\,4x-7y-4)+(-7x+14y+6,\,-4x+8y+3)\\]]>
<![CDATA[&=((8x-14y-7)+(-7x+14y+6)+1,\,(4x-7y-4)+(-4x+8y+3)+1)\\]]>
<![CDATA[&=(x,\,y)]]>
</alignmath>
</p>

<p>This final sequence of computations in $C$ is sufficient to demonstrate that any element of $C$ <em>can</em> be written (or expressed) as a linear combination of the two vectors in $R$, so $C\subseteq\spn{R}$.  Since the reverse inclusion $\spn{R}\subseteq C$ is trivially true, $C=\spn{R}$ and we say $R$ spans $C$ (<acroref type="definition" acro="SSVS" />).  Notice that this demonstration is no more or less valid if we hide from the reader our scratchwork that suggested $a_1=4x-7y-3$ and $a_2=-x+2y+1$.</p>

</example>

</subsection>

<subsection acro="VR">
<title>Vector Representation</title>

<p>In <acroref type="chapter" acro="R" /> we will take up the matter of representations fully, where <acroref type="theorem" acro="VRRB" /> will be critical for <acroref type="definition" acro="VR" />.  We will now motivate and prove a critical theorem that tells us how to <q>represent</q> a vector.   This theorem could wait, but working with it now will provide some extra insight into the nature of linearly independent spanning sets.  First an example, then the theorem.</p>

<example acro="AVR" index="vector representation">
<title>A vector representation</title>

<p>Consider the set
<equation>
S=\set{\colvector{-7\\5\\1},\,\colvector{-6\\5\\0},\,\colvector{-12\\7\\4}}
</equation>
from the vector space $\complex{3}$.  Let $A$ be the matrix whose columns are the set $S$, and verify that $A$ is nonsingular.  By <acroref type="theorem" acro="NMLIC" /> the elements of $S$ form a linearly independent set.  Suppose that $\vect{b}\in\complex{3}$.  Then $\linearsystem{A}{\vect{b}}$ has a (unique) solution (<acroref type="theorem" acro="NMUS" />) and hence is consistent.  By <acroref type="theorem" acro="SLSLC" />, $\vect{b}\in\spn{S}$.  Since $\vect{b}$ is arbitrary, this is enough to show that $\spn{S}=\complex{3}$, and therefore $S$ is a spanning set for $\complex{3}$ (<acroref type="definition" acro="SSVS" />).  (This set comes from the columns of the coefficient matrix of <acroref type="archetype" acro="B" />.)</p>

<p>Now examine the situation for a particular choice of $\vect{b}$, say $\vect{b}=\colvector{-33\\24\\5}$.  Because $S$ is a spanning set for $\complex{3}$, we know we can write $\vect{b}$ as a linear combination of the vectors in $S$,
<equation>
\colvector{-33\\24\\5}=
(-3)\colvector{-7\\5\\1}+(5)\colvector{-6\\5\\0}+(2)\colvector{-12\\7\\4}.
</equation>
</p>

<p>The nonsingularity of the matrix $A$ tells that the scalars in this linear combination are unique.  More precisely, it is the linear independence of $S$ that provides the uniqueness.  We will refer to the scalars $a_1=-3$, $a_2=5$, $a_3=2$ as a <q>representation of $\vect{b}$ relative to $S$.</q>  In other words, once we settle on $S$ as a linearly independent set that spans $\complex{3}$, the vector $\vect{b}$ is recoverable just by knowing the scalars $a_1=-3$, $a_2=5$, $a_3=2$ (use these scalars in a linear combination of the vectors in $S$).   This is all an illustration of the following important theorem, which we prove in the setting of a general vector space.</p>

</example>

<theorem acro="VRRB" index="vector representation">
<title>Vector Representation Relative to a Basis</title>
<statement>
<p>Suppose that $V$ is a vector space and $B=\set{\vectorlist{v}{m}}$ is a linearly independent set that spans $V$.  Let $\vect{w}$ be any vector in $V$.  Then there exist <em>unique</em> scalars $a_1,\,a_2,\,a_3,\,\ldots,\,a_m$ such that
<equation>
\vect{w}=\lincombo{a}{v}{m}.
</equation>
</p>

</statement>

<proof>
<p>That $\vect{w}$ can be written as a linear combination of the vectors in $B$ follows from the spanning property of the set (<acroref type="definition" acro="SSVS" />).  This is good, but not the meat of this theorem.  We now know that for any choice of the vector $\vect{w}$ there exist <em>some</em> scalars that will create $\vect{w}$ as a linear combination of the basis vectors.  The real question is:  Is there <em>more</em> than one way to write $\vect{w}$ as a linear combination of $\{\vectorlist{v}{m}\}$?  Are the scalars $a_1,\,a_2,\,a_3,\,\ldots,\,a_m$ unique?  (<acroref type="technique" acro="U" />)</p>

<p>Assume there are two different linear combinations of $\{\vectorlist{v}{m}\}$ that equal the vector $\vect{w}$.  In other words there exist scalars $a_1,\,a_2,\,a_3,\,\ldots,\,a_m$ and $b_1,\,b_2,\,b_3,\,\ldots,\,b_m$ so that
<alignmath>
<![CDATA[\vect{w}&=\lincombo{a}{v}{m}\\]]>
<![CDATA[\vect{w}&=\lincombo{b}{v}{m}.]]>
</alignmath></p>

<p>Then notice that
<alignmath>
\zerovector
<![CDATA[&=\vect{w}+(\vect{-w})&&]]>\text{<acroref type="property" acro="AI" />}\\
<![CDATA[&=\vect{w}+(-1)\vect{w}&&]]>\text{<acroref type="theorem" acro="AISM" />}\\
<![CDATA[&=(\lincombo{a}{v}{m})+\\]]>
<![CDATA[&\quad\quad(-1)(\lincombo{b}{v}{m})\\]]>
<![CDATA[&=(\lincombo{a}{v}{m})+\\]]>
<![CDATA[&\quad\quad (-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3-\ldots-b_m\vect{v}_m)&&]]>\text{<acroref type="property" acro="DVA" />}\\
<![CDATA[&=(a_1-b_1)\vect{v_1}+(a_2-b_2)\vect{v_2}+(a_3-b_3)\vect{v_3}+\\]]>
<![CDATA[&\quad\quad\cdots+(a_m-b_m)\vect{v_m}&&]]>\text{<acroref type="property" acro="C" />, <acroref type="property" acro="DSA" />}
</alignmath>
</p>

<p>But this is a relation of linear dependence on a linearly independent set of vectors (<acroref type="definition" acro="RLD" />)! Now we are using the other assumption about $B$, that $\{\vectorlist{v}{m}\}$ is a linearly independent set.   So by <acroref type="definition" acro="LI" /> it <em>must</em> happen that the scalars are all zero.  That is,
<alignmath>
<![CDATA[(a_1-b_1)&=0&(a_2-b_2)&=0&(a_3-b_3)&=0&\ldots&&(a_m-b_m)&=0\\]]>
<![CDATA[a_1&=b_1&a_2&=b_2&a_3&=b_3&\ldots&&a_m&=b_m.]]>
</alignmath>
</p>

<p>And so we find that the scalars are unique.</p>

</proof>
</theorem>

<p>The converse of <acroref type="theorem" acro="VRRB" /> is true as well, but is not important enough to rise beyond an exercise (see <acroref type="exercise" acro="LISS.T51" />).</p>

<p>This is a very typical use of the hypothesis that a set is linearly independent <mdash /> obtain a relation of linear dependence and then conclude that the scalars <em>must</em> all be zero.  The result of this theorem tells us that we can write any vector in a vector space as a linear combination of the vectors in a linearly independent spanning set, but only just.  There is only enough raw material in the spanning set to write each vector one way as a linear combination.  So in this sense, we could call a linearly independent spanning set a <q>minimal spanning set.</q>  These sets are so important that we will give them a simpler name (<q>basis</q>) and explore their properties further in the next section.</p>

</subsection>

<!--   End  liss.tex -->
<readingquestions>
<ol>
<li>Is the set of matrices below linearly independent or linearly dependent in the vector space $M_{22}$?  Why or why not?
<equation>
\set{
\begin{bmatrix}
<![CDATA[1&3\\-2&4]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[-2&3\\3&-5]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[0&9\\-1&3]]>
\end{bmatrix}
}
</equation>
</li>
<li>Explain the difference between the following two uses of the term <q>span</q>:
    <ol>
    <li>$S$ is a subset of the vector space $V$ and the span of $S$ is a subspace of $V$.</li>
    <li>$W$ is a subspace of the vector space $Y$ and $T$ spans $W$.</li>
    </ol>
</li>
<li>The set
<equation>
S=\set{
\colvector{6\\2\\1},\,
\colvector{4\\-3\\1},\,
\colvector{5\\8\\2}
}
</equation>
is linearly independent and spans $\complex{3}$.  Write the vector
$\vect{x}=\colvector{-6\\2\\2}$ as a linear combination of the elements of $S$.  How many ways are there to answer this question, and which theorem allows you to say so?
</li></ol>
</readingquestions>

<exercisesubsection>

<exercise type="C" number="20" rough="Lin ind of 3 2x2 matrices?">
<problem contributor="robertbeezer">In the vector space of $2\times 2$ matrices, $M_{22}$, determine if the set $S$ below is linearly independent.
<equation>
S=\set{
\begin{bmatrix}
<![CDATA[2&-1\\1&3]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[0&4\\-1&2]]>
\end{bmatrix},\,
\begin{bmatrix}
<![CDATA[4&2\\1&3]]>
\end{bmatrix}
}
</equation>
</problem>
<solution contributor="robertbeezer">Begin with a relation of linear dependence on the vectors in $S$ and massage it according to the definitions of vector addition and scalar multiplication in $M_{22}$,
<alignmath>
<![CDATA[\zeromatrix&=]]>
a_1
\begin{bmatrix}
<![CDATA[2&-1\\1&3]]>
\end{bmatrix}+
a_2
\begin{bmatrix}
<![CDATA[0&4\\-1&2]]>
\end{bmatrix}+
a_3
\begin{bmatrix}
<![CDATA[4&2\\1&3]]>
\end{bmatrix}\\
\begin{bmatrix}
<![CDATA[0&0\\0&0]]>
\end{bmatrix}
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[2a_1+4a_3&]]>
-a_1+4a_2+2a_3\\
<![CDATA[a_1-a_2+a_3&]]>
3a_1+2a_2+3a_3
\end{bmatrix}
</alignmath>
By our definition of matrix equality (<acroref type="definition" acro="ME" />) we arrive at a homogeneous system of linear equations,
<alignmath>
<![CDATA[2a_1+4a_3&=0\\]]>
<![CDATA[-a_1+4a_2+2a_3&=0\\]]>
<![CDATA[a_1-a_2+a_3&=0\\]]>
<![CDATA[3a_1+2a_2+3a_3&=0]]>
</alignmath>
The coefficient matrix of this system row-reduces to the matrix,
<equation>
\begin{bmatrix}
<![CDATA[\leading{1} & 0 & 0\\]]>
<![CDATA[0 & \leading{1} & 0\\]]>
<![CDATA[0 & 0 & \leading{1}\\]]>
<![CDATA[0 & 0 & 0]]>
\end{bmatrix}
</equation>
and from this we conclude that the only solution is $a_1=a_2=a_3=0$.  Since the relation of linear dependence (<acroref type="definition" acro="RLD" />) is trivial, the set $S$ is linearly independent (<acroref type="definition" acro="LI" />).
</solution>
</exercise>

<exercise type="C" number="21" rough="Lin ind of 2 crazy elements? (lin dep)">
<problem contributor="robertbeezer">In the crazy vector space $C$ (<acroref type="example" acro="CVS" />), is the set
$S=\set{(0,\,2),\ (2,\,8)}$ linearly independent?
</problem>
<solution contributor="robertbeezer">We begin with a relation of linear dependence using unknown scalars $a$ and $b$.  We wish to know if these scalars <em>must</em> both be zero.  Recall that the zero vector in $C$ is $(-1,\,-1)$ and that the definitions of vector addition and scalar multiplication are not what we might expect.
<alignmath>
\zerovector
<![CDATA[&=(-1,\,-1)\\]]>
<![CDATA[&=a(0,\,2) +b(2,\,8)&&]]>\text{<acroref type="definition" acro="RLD" />}\\
<![CDATA[&=(0a+a-1,\,2a+a-1) + (2b+b-1,\,8b+b-1)&&]]>\text{Scalar mult., <acroref type="example" acro="CVS" />}\\
<![CDATA[&=(a-1,\,3a-1) + (3b-1,\,9b-1)\\]]>
<![CDATA[&=(a-1+3b-1+1,\,3a-1+9b-1+1)&&]]>\text{Vector addition, <acroref type="example" acro="CVS" />}\\
<![CDATA[&=(a+3b-1,\,3a+9b-1)\\]]>
</alignmath>
From this we obtain two equalities, which can be converted to a homogeneous system of equations,
<alignmath>
<![CDATA[-1&=a+3b-1&a+3b&=0\\]]>
<![CDATA[-1&=3a+9b-1&3a+9b&=0]]>
</alignmath>
This homogeneous system has a singular coefficient matrix, and so has more than just the trivial solution (<acroref type="definition" acro="NM" />).  Any nontrivial solution will give us a nontrivial relation of linear dependence on $S$.  So $S$ is linearly dependent (<acroref type="definition" acro="LI" />).
</solution>
</exercise>

<exercise type="C" number="22" rough="Lin ind of 3 polynomial? (lin ind)">
<problem contributor="robertbeezer">In the vector space of polynomials $P_3$, determine if the set $S$ is linearly independent or linearly dependent.
<equation>
S=\set{2 +x -3x^2 -8x^3,\, 1+ x + x^2 +5x^3,\, 3 -4x^2 -7x^3}
</equation>
</problem>
<solution contributor="robertbeezer">Begin with a relation of linear dependence (<acroref type="definition" acro="RLD" />),
<equation>
a_1\left(2 +x -3x^2 -8x^3\right)+a_2\left(1+ x + x^2 +5x^3\right)+a_3\left(3 -4x^2 -7x^3\right)=\zerovector
</equation>
Massage according to the definitions of scalar multiplication and vector addition in the definition of $P_3$ (<acroref type="example" acro="VSP" />) and use the zero vector for this vector space,
<equation>
\left(2a_1+a_2+3a_3\right)+
\left(a_1+a_2\right)x+
\left(-3a_1+a_2-4a_3\right)x^2+
\left(-8a_1+5a_2-7a_3\right)x^3
=0+0x+0x^2+0x^3
</equation>
The definition of the equality of polynomials allows us to deduce the following four equations,
<alignmath>
<![CDATA[2a_1+a_2+3a_3&=0\\]]>
<![CDATA[a_1+a_2&=0\\]]>
<![CDATA[-3a_1+a_2-4a_3&=0\\]]>
<![CDATA[-8a_1+5a_2-7a_3&=0]]>
</alignmath>
Row-reducing the coefficient matrix of this homogeneous system leads to the unique solution $a_1=a_2=a_3=0$.  So the only relation of linear dependence on $S$ is the trivial one, and this is linear independence for $S$ (<acroref type="definition" acro="LI" />).
</solution>
</exercise>

<exercise type="C" number="23" rough="Lin ind of pair in crazy vsp? (lin dep)">
<problem contributor="robertbeezer">Determine if the set $S=\set{(3,\,1),\,(7,\,3)}$ is linearly independent in the crazy vector space $C$ (<acroref type="example" acro="CVS" />).
</problem>
<solution contributor="robertbeezer">Notice, or discover, that the following gives a nontrivial relation of linear dependence on $S$ in $C$, so by <acroref type="definition" acro="LI" />, the set $S$ is linearly dependent.
<equation>
2(3,\,1)+(-1)(7,\,3)=(7,\,3)+(-9,\,-5)=(-1,\,-1)=\zerovector
</equation>
</solution>
</exercise>

<exercise type="C" number="24" rough="dependent functions w/trig relation">
<problem contributor="chrisblack">In the vector space of real-valued functions $F = \setparts{f}{f:\mathbb{R}\rightarrow\mathbb{R}}$, determine if the following set $S$  is linearly independent.
<equation>
S = \set{\sin^2{x}, \cos^2{x}, 2}
</equation>
</problem>
<solution contributor="chrisblack">One of the fundamental identities of trigonometry is $\sin^2(x) + \cos^2(x) = 1$.  Thus, we have a dependence relation $2(\sin^2{x}) + 2(\cos^2{x}) + (-1)(2) = 0$, and the set is linearly dependent.
</solution>
</exercise>

<exercise type="C" number="25" rough="three 2x2 matrices:  independent?  span M22?">
<problem contributor="chrisblack">Let
<equation>
S = \set{
<![CDATA[\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}]]>
}
</equation>
<!-- . -->
<ol><li> Determine if $S$ spans $M_{2,2}$.
</li><li> Determine if $S$ is linearly independent.
</li></ol>
</problem>
<solution contributor="chrisblack"><ol><li>
If $S$ spans $M_{2,2}$, then for every $2 \times 2$ matrix
<![CDATA[$B = \begin{bmatrix}x & y \\ z & w\end{bmatrix}$,]]>
there exist constants $\alpha, \beta, \gamma$ so that
<alignmath>
<![CDATA[\begin{bmatrix}x & y \\ z & w\end{bmatrix}]]>
<![CDATA[&=]]>
<![CDATA[\alpha\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix} +]]>
<![CDATA[\beta\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix} +]]>
<![CDATA[\gamma\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}]]>
</alignmath>
Applying <acroref type="definition" acro="ME" />, this leads to the linear system
<alignmath>
<![CDATA[\alpha + 2\beta  &= x\\]]>
<![CDATA[2\alpha + \beta + \gamma &= y\\]]>
<![CDATA[2\alpha - \beta + \gamma &= z\\]]>
<![CDATA[\alpha + 2\beta + 2\gamma &= w.]]>
</alignmath>
We need to row-reduce the augmented matrix of this system by hand due to the symbols $x$, $y$, $z$, and $w$ in the vector of constants.
<alignmath>
\begin{bmatrix}
<![CDATA[1 & 2 & 0 & x\\]]>
<![CDATA[2 & 1 & 1 & y\\]]>
<![CDATA[2 & -1 & 1 & z\\]]>
<![CDATA[1 & 2 & 2 & w]]>
\end{bmatrix}
\rref
\begin{bmatrix}
<![CDATA[\leading{1} & 0 & 0 & x - y + z\\]]>
<![CDATA[0 & \leading{1} & 0 & \frac{1}{2}(y - z)\\]]>
<![CDATA[0 & 0 & \leading{1} & \frac{1}{2}(w - x)\\]]>
<![CDATA[0 & 0 & 0 & \frac{1}{2}(5y - 3x - 3z - w)]]>
\end{bmatrix}
</alignmath>
With the apperance of a leading 1 possible in the last column, by <acroref type="theorem" acro="RCLS" /> there will exist some matrices
<![CDATA[$B = \begin{bmatrix}x & y \\ z & w \end{bmatrix}$ so that the linear system above has no solution (namely, whenever $5y - 3x - 3z - w \ne 0$), so the set $S$ does not span $M_{2,2}$. (For example, you can verify that there is no solution when $B = \begin{bmatrix} 3 & 3 \\ 3 & 2\end{bmatrix}$.)]]>
</li><li>
To check for linear independence, we need to see if there are nontrivial coefficients $\alpha, \beta, \gamma$ that solve
<alignmath>
<![CDATA[\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}]]>
<![CDATA[&=]]>
<![CDATA[\alpha\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix} +]]>
<![CDATA[\beta\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix}+]]>
<![CDATA[\gamma\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}]]>
</alignmath>
This requires the same work that was done in part (a), with $x = y = z = w = 0$.
In that case, the coefficient matrix row-reduces to have a leading 1 in each of the first three columns and a row of zeros on the bottom, so we know that the only solution to the matrix equation is $\alpha = \beta = \gamma = 0$.  So the set $S$ is linearly independent.
</li></ol>
</solution>
</exercise>

<exercise type="C" number="26" rough="five 2x2 matrices:  independent?  span M22?">
<problem contributor="chrisblack">Let
<equation>
S = \set{
<![CDATA[\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 1 & 0\\1 & 1 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 1 & 4\\0 & 3\end{bmatrix}]]>
}
</equation>
<!-- . -->
<ol><li> Determine if $S$ spans $M_{2,2}$.
</li><li> Determine if $S$ is linearly independent.
</li></ol>
</problem>
<solution contributor="chrisblack"><ol><li>
<![CDATA[The matrices in $S$ will span $M_{2,2}$ if for any $\begin{bmatrix} x& y\\z& w\end{bmatrix}$, there are coefficients $a, b, c, d, e$ so that]]>
<alignmath>
<![CDATA[a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} +]]>
<![CDATA[b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} +]]>
<![CDATA[c\begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} +]]>
<![CDATA[d\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} +]]>
<![CDATA[e\begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix}]]>
<![CDATA[&=]]>
<![CDATA[\begin{bmatrix} x & y \\ z & w \end{bmatrix}]]>
</alignmath>
Thus, we have
<alignmath>
\begin{bmatrix}
<![CDATA[a + 2b + d + e & 2a + b + c + 4e\\]]>
<![CDATA[2a - b + c + d& a + 2b + 2c + d + 3e]]>
\end{bmatrix}
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[x & y \\]]>
<![CDATA[ z & w]]>
\end{bmatrix}\\
<intertext>so we have  the matrix equation</intertext>
\begin{bmatrix}
<![CDATA[1 & 2 & 0 & 1 & 1 \\]]>
<![CDATA[2 & 1 & 1 & 0 & 4\\]]>
<![CDATA[2 & -1 & 1 & 1 & 0\\]]>
<![CDATA[1 & 2 & 2 & 1 & 3]]>
\end{bmatrix}
\colvector{a\\b\\c\\d\\e}
<![CDATA[&=]]>
\colvector{x\\y\\z\\w}
</alignmath>
This system will have a solution for <em>every</em> vector on the right side if the row-reduced coefficient matrix has a leading one in every row, since then it is never possible to have a leading 1 appear in the final column of a row-reduced augmented matrix.
<alignmath>
\begin{bmatrix}
<![CDATA[1 & 2 & 0 & 1 & 1 \\]]>
<![CDATA[2 & 1 & 1 & 0 & 4\\]]>
<![CDATA[2 & -1 & 1 & 1 & 0\\]]>
<![CDATA[1 & 2 & 2 & 1 & 3]]>
\end{bmatrix}
\rref
\begin{bmatrix}
<![CDATA[\leading{1}& 0 & 0 &  0 & 1\\]]>
<![CDATA[0 & \leading{1} & 0 & 0 & 1\\]]>
<![CDATA[0 & 0 & \leading{1} & 0 & 1\\]]>
<![CDATA[0 &  0 & 0 & \leading{1} & -2]]>
\end{bmatrix}
</alignmath>
<![CDATA[Since there is a leading one in each row of the row-reduced coefficient matrix, there is a solution for every vector $\colvector{x\\y\\z\\w}$, which means that there is a solution to the original equation for every matrix $\begin{bmatrix} x & y\\ z & w \end{bmatrix}$.  Thus, the original five matrices span $M_{2,2}$.]]>
</li><li> The matrices in $S$ are linearly independent if the only solution to
<alignmath>
<![CDATA[a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} +]]>
<![CDATA[b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} +]]>
<![CDATA[c\begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} +]]>
<![CDATA[d\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} +]]>
<![CDATA[e\begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix}]]>
<![CDATA[&=]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}]]>
</alignmath>
is $a = b = c = d = e = 0$.<br /><br />
We have
<alignmath>
\begin{bmatrix}
<![CDATA[a + 2b + d + e & 2a + b + c + 4e\\]]>
<![CDATA[2a - b + c + d & a + 2b + 2c + d + 3e]]>
\end{bmatrix}
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[1 & 2 & 0 & 1 & 1 \\]]>
<![CDATA[2 & 1 & 1 & 0 & 4\\]]>
<![CDATA[2 & -1 & 1 & 1 & 0\\]]>
<![CDATA[1 & 2 & 2 & 1 & 3 \end{bmatrix}]]>
\colvector{a\\b\\c\\d\\e}
=
\begin{bmatrix}
<![CDATA[0 & 0 \\]]>
<![CDATA[0 & 0 \end{bmatrix}]]>
</alignmath>
so we need to find the nullspace of the matrix
<alignmath>
\begin{bmatrix}
<![CDATA[1 & 2 & 0 & 1 & 1 \\]]>
<![CDATA[2 & 1 & 1 & 0 & 4\\]]>
<![CDATA[2 & -1 & 1 & 1 & 0\\]]>
<![CDATA[1 & 2 & 2 & 1 & 3]]>
\end{bmatrix}
</alignmath>
We row-reduced this matrix in part (a), and found that there is a column without a leading 1, which correspons to a free variable in a description of the solution set to the homogeneous system, so the nullspace is nontrivial and there are an infinite number of solutions to
<alignmath>
<![CDATA[a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} +]]>
<![CDATA[b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} +]]>
<![CDATA[c \begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} +]]>
<![CDATA[d \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} +]]>
<![CDATA[e \begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix}]]>
<![CDATA[&=]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}]]>
</alignmath>
Thus, this set of matrices is not linearly independent.
</li></ol>
</solution>
</exercise>

<exercise type="C" number="30" rough="RLD in Example LIM32">
<problem contributor="robertbeezer">In <acroref type="example" acro="LIM32" />, find another nontrivial relation of linear dependence on the linearly dependent set of $3\times 2$ matrices, $S$.
</problem>
</exercise>

<exercise type="C" number="40" rough="3 vectors in P_4 do not span">
<problem contributor="robertbeezer">Determine if the set $T=\set{x^2-x+5,\,4x^3-x^2+5x,\,3x+2}$ spans the vector space of polynomials with degree 4 or less, $P_4$.
</problem>
<solution contributor="robertbeezer">The polynomial $x^4$ is an element of $P_4$.  Can we write this element as a linear combination of the elements of $T$?  To wit, are there scalars $a_1$, $a_2$, $a_3$ such that
<alignmath>
<![CDATA[x^4&=a_1\left(x^2-x+5\right)+a_2\left(4x^3-x^2+5x\right)+a_3\left(3x+2\right)]]>
</alignmath>
Massaging the right side of this equation, according to the definitions of <acroref type="example" acro="VSP" />, and then equating coefficients, leads to an inconsistent system of equations (check this!).  As such, $T$ is not a spanning set for $P_4$.
</solution>
</exercise>

<exercise type="C" number="41" rough="4 matrices in subspace do span">
<problem contributor="robertbeezer">The set $W$ is a subspace of $M_{22}$, the vector space of all $2\times 2$ matrices.  Prove that $S$ is a spanning set for $W$.
<alignmath>
<![CDATA[W&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{2a-3b+4c-d=0}]]>
<![CDATA[&]]>
<![CDATA[S=&\set{]]>
<![CDATA[\begin{bmatrix}1&0\\0&2\end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix}0&1\\0&-3\end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix}0&0\\1&4\end{bmatrix}]]>
}
</alignmath>
</problem>
<solution contributor="robertbeezer">We want to show that $W=\spn{S}$ (<acroref type="definition" acro="SSVS" />), which is an equality of sets (<acroref type="definition" acro="SE" />).<br /><br />
First, show that $\spn{S}\subseteq W$.  Begin by checking that each of the three matrices in $S$ is a member of the set $W$.  Then, since $W$ is a vector space, the closure properties (<acroref type="property" acro="AC" />, <acroref type="property" acro="SC" />) guarantee that every linear combination of elements of $S$ remains in $W$.<br /><br />
Second, show that $W\subseteq\spn{S}$.    We want to convince ourselves that an arbitrary element of $W$ is a linear combination of elements of $S$.  Choose
<equation>
<![CDATA[\vect{x}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\in W]]>
</equation>
The values of $a,\,b,\,c,\,d$ are not totally arbitrary, since membership in $W$ requires that
$2a-3b+4c-d=0$.  Now, rewrite as follows,
<alignmath>
\vect{x}
<![CDATA[&=\begin{bmatrix}a&b\\c&d\end{bmatrix}\\]]>
<![CDATA[&=\begin{bmatrix}a&b\\c&2a-3b+4c\end{bmatrix}]]>
<![CDATA[&&2a-3b+4c-d=0\\]]>
<![CDATA[&=]]>
<![CDATA[\begin{bmatrix}a&0\\0&2a\end{bmatrix}+]]>
<![CDATA[\begin{bmatrix}0&b\\0&-3b\end{bmatrix}+]]>
<![CDATA[\begin{bmatrix}0&0\\c&4c\end{bmatrix}]]>
<![CDATA[&&]]>\text{<acroref type="definition" acro="MA" />}\\
<![CDATA[&=]]>
<![CDATA[a\begin{bmatrix}1&0\\0&2\end{bmatrix}+]]>
<![CDATA[b\begin{bmatrix}0&1\\0&-3\end{bmatrix}+]]>
<![CDATA[c\begin{bmatrix}0&0\\1&4\end{bmatrix}]]>
<![CDATA[&&]]>\text{<acroref type="definition" acro="MSM" />}\\
<![CDATA[&\in\spn{S}]]>
<![CDATA[&&]]>\text{<acroref type="definition" acro="SS" />}
</alignmath>
</solution>
</exercise>

<exercise type="C" number="42" rough="Pair in crazy vsp spans? (no)">
<problem contributor="robertbeezer">Determine if the set $S=\set{(3,\,1),\,(7,\,3)}$ spans the crazy vector space $C$ (<acroref type="example" acro="CVS" />).
</problem>
<solution contributor="robertbeezer">We will try to show that $S$ spans $C$.  Let $(x,\,y)$ be an arbitrary element of $C$ and search for scalars $a_1$ and $a_2$ such that
<alignmath>
<![CDATA[(x,\,y)&=a_1(3,\,1) + a_2(7,\,3)\\]]>
<![CDATA[&=(4a_1-1,\,2a_1-1)+(8a_2-1,\,4a_2-1)\\]]>
<![CDATA[&=(4a_1+8a_2-1,2a_1+4a_2-1)]]>
</alignmath>
Equality in $C$ leads to the system
<alignmath>
<![CDATA[4a_1+8a_2&=x+1\\]]>
<![CDATA[2a_1+4a_2&=y+1]]>
</alignmath>
This system has a singular coefficient matrix whose column space is simply $\spn{\colvector{2\\1}}$.  So any choice of $x$ and $y$ that causes the column vector $\colvector{x+1\\y+1}$ to lie outside the column space will lead to an inconsistent system, and hence create an element $(x,\,y)$ that is not in the span of $S$.  So $S$ does not span $C$.<br /><br />
For example, choose $x=0$ and $y=5$, and then we can see that $\colvector{1\\6}\not\in\spn{\colvector{2\\1}}$ and we know that $(0,\,5)$ cannot be written as a linear combination of the vectors in $S$.  A shorter solution might begin by asserting that $(0,\,5)$ is not in $\spn{S}$ and then establishing this claim alone.
</solution>
</exercise>

<exercise type="M" number="10" rough="Example SSP4 with RNS">
<problem contributor="robertbeezer">Halfway through <acroref type="example" acro="SSP4" />, we need to show that the system of equations
<equation>
\linearsystem{
\begin{bmatrix}
<![CDATA[0 & 0 & 0 & 1\\]]>
<![CDATA[0 & 0 & 1 & -8\\]]>
<![CDATA[0 & 1 & -6 & 24\\]]>
<![CDATA[1 & -4 & 12 & -32\\]]>
<![CDATA[-2 & 4 & -8 & 16\\]]>
\end{bmatrix}
}
{\colvector{a\\b\\c\\d\\e}}
</equation>
is consistent for every choice of the vector of constants satisfying $16a+8b+4c+2d+e=0$.<br /><br />
Express the column space of the coefficient matrix of this system as a null space, using <acroref type="theorem" acro="FS" />.  From this use <acroref type="theorem" acro="CSCS" /> to establish that the system is always consistent.  Notice that this approach removes from <acroref type="example" acro="SSP4" /> the need to row-reduce a symbolic matrix.
</problem>
<solution contributor="robertbeezer"><acroref type="theorem" acro="FS" /> provides the matrix
<equation>
L=
\begin{bmatrix}
<![CDATA[\leading{1} & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16}]]>
\end{bmatrix}
</equation>
and so if $A$ denotes the coefficient matrix of the system, then $\csp{A}=\nsp{L}$.  The single homogeneous equation in $\homosystem{L}$ is equivalent to the condition on the vector of constants (use $a,\,b,\,c,\,d,\,e$ as variables and then multiply by 16).
</solution>
</exercise>

<exercise type="T" number="20" rough="subset of linearly independent is linearly independent">
<problem contributor="robertbeezer">Suppose that $S$ is a finite linearly independent set of vectors from the vector space $V$.  Let $T$ be any subset of $S$.  Prove that $T$ is linearly independent.
</problem>
<solution contributor="robertbeezer">We will prove the contrapositive (<acroref type="technique" acro="CP" />): If $T$ is linearly dependent, then $S$ is linearly dependent.  This might be an interesting statement in its own right.<br /><br />
Write $S=\set{\vectorlist{v}{m}}$ and without loss of generality we can assume that the subset $T$ is the first $t$ vectors of $S$, $t\leq m$, so $T=\set{\vectorlist{v}{t}}$.  Since $T$ is linearly dependent, by <acroref type="definition" acro="LI" /> there are scalars, not all zero, $\scalarlist{a}{t}$, so that
<alignmath>
\zerovector
<![CDATA[&= \lincombo{a}{v}{t}\\]]>
<![CDATA[&= \lincombo{a}{v}{t}+\zerovector+\zerovector+\dots+\zerovector\\]]>
<![CDATA[&= \lincombo{a}{v}{t}+0\vect{v}_{t+1}+0\vect{v}_{t+2}+\dots+0\vect{v}_{m}]]>
</alignmath>
which is a nontrivial relation of linear dependence (<acroref type="definition" acro="RLD" />) on the set $S$, so we can say $S$ is linearly dependent.
</solution>
</exercise>

<exercise type="T" number="40" rough="Generalize Theorem EMMVP to bases, spanning sets">
<problem contributor="robertbeezer">Prove the following variant of <acroref type="theorem" acro="EMMVP" /> that has a weaker hypothesis:  Suppose that $C=\set{\vectorlist{u}{p}}$ is a linearly independent spanning set for $\complex{n}$.  Suppose also that $A$ and $B$ are $m\times n$ matrices such that $A\vect{u}_i=B\vect{u}_i$ for every $1\leq i\leq n$.  Then $A=B$.<br /><br />
Can you weaken the hypothesis even further while still preserving the conclusion?
</problem>
</exercise>

<exercise type="T" number="50" rough="Linear dependence in a set of two vectors">
<problem contributor="robertbeezer">Suppose that $V$ is a vector space and $\vect{u},\,\vect{v}\in V$ are two vectors in $V$.  Use the definition of linear independence to prove that $S=\set{\vect{u},\,\vect{v}}$ is a linearly dependent set if and only if one of the two vectors is a scalar multiple of the other.  Prove this directly in the context of an abstract vector space ($V$), without simply giving an upgraded version of <acroref type="theorem" acro="DLDS" /> for the special case of just two vectors.
</problem>
<solution contributor="robertbeezer"><implyforward /> If $S$ is linearly dependent, then there are scalars $\alpha$ and $\beta$, not both zero, such that $\alpha\vect{u}+\beta\vect{v}=\zerovector$.  Suppose that $\alpha\neq 0$, the proof proceeds similarly if $\beta\neq 0$.  Now,
<alignmath>
\vect{u}
<![CDATA[&=1\vect{u}&&]]>\text{<acroref type="property" acro="O" />}\\
<![CDATA[&=\left(\frac{1}{\alpha}\alpha\right)\vect{u}&&]]>\text{<acroref type="property" acro="MICN" />}\\
<![CDATA[&=\frac{1}{\alpha}\left(\alpha\vect{u}\right)&&]]>\text{<acroref type="property" acro="SMA" />}\\
<![CDATA[&=\frac{1}{\alpha}\left(\alpha\vect{u}+\zerovector\right)&&]]>\text{<acroref type="property" acro="Z" />}\\
<![CDATA[&=\frac{1}{\alpha}\left(\alpha\vect{u}+\beta\vect{v}-\beta\vect{v}\right)&&]]>\text{<acroref type="property" acro="AI" />}\\
<![CDATA[&=\frac{1}{\alpha}\left(\zerovector-\beta\vect{v}\right)&&]]>\text{<acroref type="definition" acro="LI" />}\\
<![CDATA[&=\frac{1}{\alpha}\left(-\beta\vect{v}\right)&&]]>\text{<acroref type="property" acro="Z" />}\\
<![CDATA[&=\frac{-\beta}{\alpha}\vect{v}&&]]>\text{<acroref type="property" acro="SMA" />}
</alignmath>
which shows that $\vect{u}$ is a scalar multiple of $\vect{v}$.<br /><br />
<implyreverse /> Suppose now that $\vect{u}$ is a scalar multiple of $\vect{v}$.  More precisely, suppose there is a scalar $\gamma$ such that $\vect{u}=\gamma\vect{v}$.  Then
<alignmath>
(-1)\vect{u}+\gamma\vect{v}
<![CDATA[&=(-1)\vect{u}+\vect{u}\\]]>
<![CDATA[&=(-1)\vect{u}+(1)\vect{u}&&]]>\text{<acroref type="property" acro="O" />}\\
<![CDATA[&=\left((-1)+1\right)\vect{u}&&]]>\text{<acroref type="property" acro="DSA" />}\\
<![CDATA[&=0\vect{u}&&]]>\text{<acroref type="property" acro="AICN" />}\\
<![CDATA[&=\zerovector&&]]>\text{<acroref type="theorem" acro="ZSSM" />}
</alignmath>
This is a relation of linear of linear dependence on $S$ (<acroref type="definition" acro="RLD" />), which is nontrivial since one of the scalars is $-1$.  Therefore $S$ is linearly dependent by <acroref type="definition" acro="LI" />.<br /><br />
Be careful using this theorem.  It is only applicable to sets of two vectors.  In particular, linear dependence in a set of three or more vectors can be more complicated than just one vector being a scalar multiple of another.
</solution>
</exercise>

<exercise type="T" number="51" rough="converse of vector representation">
<problem contributor="robertbeezer">Carefully formulate the converse of <acroref type="theorem" acro="VRRB" /> and provide a proof.
</problem>
<solution contributor="robertbeezer">The converse could read: <q>Suppose that $V$ is a vector space and $S=\set{\vectorlist{v}{m}}$ is a set of vectors in $V$.  If, for each $\vect{w}\in V$, there are <em>unique</em> scalars $a_1,\,a_2,\,a_3,\,\ldots,\,a_m$ such that
<equation>
\vect{w}=\lincombo{a}{v}{m}
</equation>
then $S$ is a linearly independent set that spans $V$.</q><br /><br />
Since every vector $\vect{w}\in V$ is assumed to be a linear combination of the elements of $S$, it is easy to see that $S$ is a spanning set for $V$ (<acroref type="definition" acro="SSVS" />).<br /><br />
To establish linear independence, begin with an arbitrary relation of linear dependence on the vectors in $S$ (<acroref type="definition" acro="RLD" />).  One way to form such a relation is the trivial way, where each scalar is zero.   But our hypothesis of uniqueness then implies that that the <em>only</em> way to form this relation of linear dependence is the trivial way.  But this establishes the linear independence of $S$ (<acroref type="definition" acro="LI" />).
</solution>
</exercise>

</exercisesubsection>

</section>