Source

fcla / src / section-SLT.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
1159
1160
1161
1162
1163
1164
1165
1166
1167
1168
1169
1170
1171
1172
1173
1174
1175
1176
1177
1178
1179
1180
1181
1182
1183
1184
1185
1186
1187
1188
1189
1190
1191
1192
1193
1194
1195
1196
1197
1198
1199
1200
1201
1202
1203
1204
1205
1206
1207
1208
1209
1210
1211
1212
1213
1214
1215
1216
1217
1218
1219
1220
1221
1222
1223
1224
1225
1226
1227
1228
1229
1230
1231
1232
1233
1234
1235
1236
1237
1238
1239
1240
1241
1242
1243
1244
1245
1246
1247
1248
1249
1250
1251
1252
1253
1254
1255
1256
1257
1258
1259
1260
1261
1262
1263
1264
1265
1266
1267
1268
1269
1270
1271
1272
1273
1274
1275
1276
1277
1278
1279
1280
1281
1282
1283
1284
1285
1286
1287
1288
1289
1290
<?xml version="1.0" encoding="UTF-8" ?>
<section acro="SLT">
<title>Surjective Linear Transformations</title>

<!-- %%%%%%%%%% -->
<!-- % -->
<!-- %  Section SLT -->
<!-- %  Surjective Linear Transformations -->
<!-- % -->
<!-- %%%%%%%%%% -->
<introduction>
<p>The companion to an injection is a surjection.  Surjective linear transformations are closely related to spanning sets and ranges.  So as you read this section reflect back on <acroref type="section" acro="ILT" /> and note the parallels and the contrasts.  In the next section, <acroref type="section" acro="IVLT" />, we will combine the two properties.</p>

</introduction>

<subsection acro="SLT">
<title>Surjective Linear Transformations</title>

<p>As usual, we lead with a definition.</p>

<definition acro="SLT" index="linear transformation!surjection">
<title>Surjective Linear Transformation</title>
<p>Suppose $\ltdefn{T}{U}{V}$ is a linear transformation.  Then $T$ is <define>surjective</define> if for every $\vect{v}\in V$ there exists a $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$.</p>

</definition>

<p>Given an arbitrary function, it is possible for there to be an element of the codomain that is not an output of the function (think about the function $y=f(x)=x^2$ and the codomain element $y=-3$).  For a surjective function, this never happens.  If we choose any element of the codomain ($\vect{v}\in V$) then there must be an input from the domain ($\vect{u}\in U$) which will create the output when used to evaluate the linear transformation ($\lt{T}{\vect{u}}=\vect{v}$).  Some authors prefer the term <define>onto</define> where we use surjective, and we will sometimes refer to a surjective linear transformation as a <define>surjection</define>.</p>

</subsection>

<subsection acro="ESLT">
<title>Examples of Surjective Linear Transformations</title>

<p>It is perhaps most instructive to examine a linear transformation that is not surjective first.</p>

<example acro="NSAQ" index="surjective!not">
<title>Not surjective, Archetype Q</title>

<p><acroref type="archetype" acro="Q" /> is the linear transformation
<equation>
<archetypepart acro="Q" part="ltdefn" /></equation></p>

<p>We will demonstrate that
<equation>
\vect{v}=\colvector{-1\\2\\3\\-1\\4}
</equation>
is an unobtainable element of the codomain.  Suppose to the contrary that $\vect{u}$ is an element of the domain such that $\lt{T}{\vect{u}}=\vect{v}$.</p>

<p>Then
<alignmath>
<![CDATA[\colvector{-1\\2\\3\\-1\\4}&=\vect{v}=\lt{T}{\vect{u}}]]>
=\lt{T}{\colvector{u_1\\u_2\\u_3\\u_4\\u_5}}\\
<![CDATA[&=\colvector{]]>
-2 u_1 + 3 u_2 + 3 u_3 - 6 u_4 + 3 u_5\\
-16 u_1 + 9 u_2 + 12 u_3 - 28 u_4 + 28 u_5\\
-19 u_1 + 7 u_2 + 14 u_3 - 32 u_4 + 37 u_5\\
-21 u_1 + 9 u_2 + 15 u_3 - 35 u_4 + 39 u_5\\
-9 u_1 + 5 u_2 + 7 u_3 - 16 u_4 + 16 u_5}\\
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[-2&3&3&-6&3\\]]>
<![CDATA[-16&9&12&-28&28\\]]>
<![CDATA[-19&7&14&-32&37\\]]>
<![CDATA[-21&9&15&-35&39\\]]>
<![CDATA[-9&5&7&-16&16]]>
\end{bmatrix}
\colvector{u_1\\u_2\\u_3\\u_4\\u_5}
</alignmath>
</p>

<p>Now we recognize the appropriate input vector $\vect{u}$ as a solution to a linear system of equations.  Form the augmented matrix of the system, and row-reduce to
<equation>
\begin{bmatrix}
<![CDATA[\leading{1} & 0 & 0 & 0 & -1 & 0\\]]>
<![CDATA[0 & \leading{1} & 0 & 0 & -\frac{4}{3} & 0\\]]>
<![CDATA[0 & 0 & \leading{1} & 0 & -\frac{1}{3} & 0\\]]>
<![CDATA[0 & 0 & 0 & \leading{1} & -1 & 0\\]]>
<![CDATA[0 & 0 & 0 & 0 & 0 & \leading{1}\\]]>
\end{bmatrix}
</equation>
</p>

<p>With a leading 1 in the last column, <acroref type="theorem" acro="RCLS" /> tells us the system is inconsistent.  From the absence of any solutions we conclude that no such vector $\vect{u}$ exists, and by <acroref type="definition" acro="SLT" />, $T$ is not surjective.</p>

<p>Again, do not concern yourself with how $\vect{v}$ was selected, as this will be explained shortly.  However, do understand <em>why</em> this vector provides enough evidence to conclude that $T$ is not surjective.</p>
</example>

<p>Here's a cartoon of a non-surjective linear transformation.  Notice that the central feature of this cartoon is that the vector $\vect{v}\in V$ does not have an arrow pointing to it, implying there is no $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$.  Even though this happens again with a second unnamed vector in $V$, it only takes one occurrence to destroy the possibility of surjectivity.
<diagram acro="NSLT">
<title>Non-Surjective Linear Transformation</title>
<tikz>
\tikzset{ltvect/.style={shape=circle, minimum size=0.30em, inner sep=0pt, draw, fill=black}}
<![CDATA[\tikzset{ltedge/.style={->, bend left=20, thick, shorten <=0.1em, shorten >=0.1em}}]]>
% base generic picture
\draw ( 5em, 8em) circle [x radius=5em, y radius=8em, thick];
\draw (20em, 8em) circle [x radius=5em, y radius=8em, thick];
\node (U) at ( 5em, -1em) {$U$};
\node (V) at (20em, -1em) {$V$};
\draw[->, thick, draw] (U) to node[auto] {$T$} (V);
% inputs
\node (u1) [ltvect]                         at (5em, 13em) {};
\node (u2) [ltvect]                         at (5em, 11em) {};
\node (u3) [ltvect]                         at (5em,  8em) {};
\node (u4) [ltvect]                         at (5em,  6em) {};
% outputs
\node (v1) [ltvect]                         at (20em, 12em) {};
\node (v2) [ltvect]                         at (20em,  7em) {};
\node (v3) [ltvect]                         at (19em,  3em) {};
\node (v)  [ltvect, label=right:$\vect{v}$] at (21em,  3em) {};
% associations
\draw[ltedge] (u1) to (v1);
\draw[ltedge] (u2) to (v1);
\draw[ltedge] (u3) to (v2);
\draw[ltedge] (u4) to (v2);
</tikz>
</diagram>
</p>

<p>To show that a linear transformation is not surjective, it is enough to find a single element of the codomain that is never created by any input, as in <acroref type="example" acro="NSAQ" />.  However, to show that a linear transformation is surjective we must establish that <em>every</em> element of the codomain occurs as an output of the linear transformation for some appropriate input.</p>

<example acro="SAR" index="surjective">
<title>Surjective, Archetype R</title>

<p><acroref type="archetype" acro="R" /> is the linear transformation
<equation>
<archetypepart acro="R" part="ltdefn" /></equation>
</p>

<p>To establish that $R$ is surjective we must begin with a totally arbitrary element of the codomain, $\vect{v}$ and somehow find an input vector $\vect{u}$ such that $\lt{T}{\vect{u}}=\vect{v}$.  We desire,
<alignmath>
<![CDATA[\lt{T}{\vect{u}}&=\vect{v}\\]]>
\colvector{-65 u_1 + 128 u_2 + 10 u_3 - 262 u_4 + 40 u_5\\
36 u_1 - 73 u_2 - u_3 + 151 u_4 - 16 u_5\\
-44 u_1 + 88 u_2 + 5 u_3 - 180 u_4 + 24 u_5\\
34 u_1 - 68 u_2 - 3 u_3 + 140 u_4 - 18 u_5\\
12 u_1 - 24 u_2 - u_3 + 49 u_4 - 5 u_5}
<![CDATA[&=]]>
\colvector{v_1\\v_2\\v_3\\v_4\\v_5}\\
\begin{bmatrix}
<![CDATA[-65&128&10&-262&40\\]]>
<![CDATA[36&-73&-1&151&-16\\]]>
<![CDATA[-44&88&5&-180&24\\]]>
<![CDATA[34&-68&-3&140&-18\\]]>
<![CDATA[12&-24&-1&49&-5]]>
\end{bmatrix}
\colvector{u_1\\u_2\\u_3\\u_4\\u_5}
<![CDATA[&=]]>
\colvector{v_1\\v_2\\v_3\\v_4\\v_5}
</alignmath>
</p>

<p>We recognize this equation as a system of equations in the variables $u_i$, but our vector of constants contains symbols.  In general, we would have to row-reduce the augmented matrix by hand, due to the symbolic final column.  However, in this particular example, the $5\times 5$ coefficient matrix is nonsingular and so has an inverse (<acroref type="theorem" acro="NI" />, <acroref type="definition" acro="MI" />).
<equation>
\inverse{
\begin{bmatrix}
<![CDATA[-65&128&10&-262&40\\]]>
<![CDATA[36&-73&-1&151&-16\\]]>
<![CDATA[-44&88&5&-180&24\\]]>
<![CDATA[34&-68&-3&140&-18\\]]>
<![CDATA[12&-24&-1&49&-5]]>
\end{bmatrix}
}
=
\begin{bmatrix}
<![CDATA[-47 & 92 &  1 & -181 & -14 \\]]>
<![CDATA[ 27 & -55 & \frac{7}{2} & \frac{221}{2} & 11\\]]>
<![CDATA[-32 & 64  & -1 &  -126 &  -12\\]]>
<![CDATA[ 25 &  -50 &  \frac{3}{2} & \frac{199}{2} & 9 \\]]>
<![CDATA[ 9 & -18 & \frac{1}{2} & \frac{71}{2} & 4]]>
\end{bmatrix}
</equation>
so we find that
<alignmath>
\colvector{u_1\\u_2\\u_3\\u_4\\u_5}
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[-47 & 92 &  1 & -181 & -14 \\]]>
<![CDATA[ 27 & -55 & \frac{7}{2} & \frac{221}{2} & 11\\]]>
<![CDATA[-32 & 64  & -1 &  -126 &  -12\\]]>
<![CDATA[ 25 &  -50 &  \frac{3}{2} & \frac{199}{2} & 9 \\]]>
<![CDATA[ 9 & -18 & \frac{1}{2} & \frac{71}{2} & 4]]>
\end{bmatrix}
\colvector{v_1\\v_2\\v_3\\v_4\\v_5}\\
<![CDATA[&=]]>
\colvector{-47 v_1 + 92 v_2 + v_3 - 181 v_4 - 14 v_5\\
 27 v_1 - 55 v_2 + \frac{7}{2} v_3 + \frac{221}{2} v_4  + 11 v_5\\
-32 v_1 + 64  v_2 - v_3 - 126 v_4 - 12 v_5\\
 25 v_1 - 50 v_2 + \frac{3}{2} v_3 + \frac{199}{2} v_4 + 9 v_5\\
 9 v_1 - 18 v_2 + \frac{1}{2} v_3 + \frac{71}{2} v_4 + 4 v_5}
</alignmath>
</p>

<p>This establishes that if we are given <em>any</em> output vector $\vect{v}$, we can use its components in this final expression to formulate a vector $\vect{u}$ such that $\lt{T}{\vect{u}}=\vect{v}$.  So by <acroref type="definition" acro="SLT" /> we now know that $T$ is surjective.  You might try to verify this condition in its full generality (<ie /> evaluate $T$ with this final expression and see if you get $\vect{v}$ as the result), or test it more specifically for some numerical vector $\vect{v}$ (see <acroref type="exercise" acro="SLT.C20" />).</p>
</example>

<p>Here's the cartoon for a surjective linear transformation.  It is meant to suggest that for every output in $V$ there is <em>at least one</em> input in $U$ that is sent to the output.  (Even though we have depicted several inputs sent to each output.)  The key feature of this cartoon is that there are no vectors in $V$ without an arrow pointing to them.
<diagram acro="SLT">
<title>Surjective Linear Transformation</title>
<tikz>
\tikzset{ltvect/.style={shape=circle, minimum size=0.30em, inner sep=0pt, draw, fill=black}}
<![CDATA[\tikzset{ltedge/.style={->, bend left=20, thick, shorten <=0.1em, shorten >=0.1em}}]]>
% base generic picture
\draw ( 5em, 8em) circle [x radius=5em, y radius=8em, thick];
\draw (20em, 8em) circle [x radius=4em, y radius=7em, thick];
\node (U) at ( 5em, -1em) {$U$};
\node (V) at (20em, -1em) {$V$};
\draw[->, thick, draw] (U) to node[auto] {$T$} (V);
% inputs
\node (u1) [ltvect]                         at (5em, 14em) {};
\node (u2) [ltvect]                         at (5em, 12em) {};
\node (u3) [ltvect]                         at (5em,  9em) {};
\node (u4) [ltvect]                         at (5em,  7em) {};
\node (u5) [ltvect]                         at (5em,  4em) {};
\node (u6) [ltvect]                         at (5em,  2em) {};
% outputs
\node (v1) [ltvect]                         at (20em, 12em) {};
\node (v2) [ltvect]                         at (20em,  8em) {};
\node (v3) [ltvect]                         at (20em,  4em) {};
% associations
\draw[ltedge] (u1) to (v1);
\draw[ltedge] (u2) to (v1);
\draw[ltedge] (u3) to (v2);
\draw[ltedge] (u4) to (v2);
\draw[ltedge] (u5) to (v3);
\draw[ltedge] (u6) to (v3);
</tikz>
</diagram>
</p>

<p>Let's now examine a surjective linear transformation between abstract vector spaces.</p>

<example acro="SAV" index="surjective!polynomials to matrices">
<title>Surjective, Archetype V</title>

<p><acroref type="archetype" acro="V" /> is defined by
<equation>
<archetypepart acro="V" part="ltdefn" /></equation>
</p>

<p>To establish that the linear transformation is surjective, begin by choosing an arbitrary output.  In this example, we need to choose an arbitrary $2\times 2$ matrix, say
<equation>
<![CDATA[\vect{v}=\begin{bmatrix}x&y\\z&w\end{bmatrix}]]>
</equation>
and we would like to find an input polynomial
<equation>
\vect{u}=a+bx+cx^2+dx^3
</equation>
so that $\lt{T}{\vect{u}}=\vect{v}$.  So we have,
<alignmath>
<![CDATA[\begin{bmatrix}x&y\\z&w\end{bmatrix}&=\vect{v}\\]]>
<![CDATA[&=\lt{T}{\vect{u}}\\]]>
<![CDATA[&=\lt{T}{a+bx+cx^2+dx^3}\\]]>
<![CDATA[&=\begin{bmatrix}]]>
<![CDATA[a+b & a-2c\\]]>
<![CDATA[d & b-d]]>
\end{bmatrix}
</alignmath>
</p>

<p>Matrix equality leads us to the system of four equations in the four unknowns, $x,y,z,w$,
<alignmath>
<![CDATA[a+b&=x\\]]>
<![CDATA[a-2c&=y\\]]>
<![CDATA[d&=z\\]]>
<![CDATA[b-d&=w]]>
</alignmath>
which can be rewritten as a matrix equation,
<equation>
\begin{bmatrix}
<![CDATA[1 & 1 & 0 & 0\\]]>
<![CDATA[1 & 0 & -2 & 0 \\]]>
<![CDATA[0 & 0 & 0 & 1\\]]>
<![CDATA[0 & 1 & 0 & -1]]>
\end{bmatrix}
\colvector{a\\b\\c\\d}
=
\colvector{x\\y\\z\\w}
</equation>
</p>

<p>The coefficient matrix is nonsingular, hence it has an inverse,
<equation>
\inverse{
\begin{bmatrix}
<![CDATA[1 & 1 & 0 & 0\\]]>
<![CDATA[1 & 0 & -2 & 0 \\]]>
<![CDATA[0 & 0 & 0 & 1\\]]>
<![CDATA[0 & 1 & 0 & -1]]>
\end{bmatrix}
}
=
\begin{bmatrix}
<![CDATA[1 & 0 & -1 & -1\\]]>
<![CDATA[0 & 0 & 1 & 1\\]]>
<![CDATA[\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\]]>
<![CDATA[0 & 0 & 1 & 0]]>
\end{bmatrix}
</equation>
so we have
<alignmath>
<![CDATA[\colvector{a\\b\\c\\d}&=]]>
\begin{bmatrix}
<![CDATA[1 & 0 & -1 & -1\\]]>
<![CDATA[0 & 0 & 1 & 1\\]]>
<![CDATA[\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\]]>
<![CDATA[0 & 0 & 1 & 0]]>
\end{bmatrix}
\colvector{x\\y\\z\\w}\\%
<![CDATA[&=]]>
\colvector{
x-z-w\\
z+w\\
\frac{1}{2}(x-y-z-w)\\
z
}
</alignmath>
</p>

<p>So the input polynomial $\vect{u}=(x-z-w)+(z+w)x+\frac{1}{2}(x-y-z-w)x^2+zx^3$ will yield the output matrix $\vect{v}$, no matter what form $\vect{v}$ takes.  This means by <acroref type="definition" acro="SLT" /> that $T$ is surjective.  All the same, let's do a concrete demonstration and evaluate $T$ with $\vect{u}$,
<alignmath>
<![CDATA[\lt{T}{\vect{u}}&=\lt{T}{(x-z-w)+(z+w)x+\frac{1}{2}(x-y-z-w)x^2+zx^3}\\]]>
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[(x-z-w)+(z+w) & (x-z-w) - 2(\frac{1}{2}(x-y-z-w))\\]]>
<![CDATA[z & (z+w)-z]]>
\end{bmatrix}\\
<![CDATA[&=]]>
\begin{bmatrix}
<![CDATA[x & y\\]]>
<![CDATA[z & w]]>
\end{bmatrix}\\
<![CDATA[&=\vect{v}]]>
</alignmath>
</p>

</example>

</subsection>

<subsection acro="RLT">
<title>Range of a Linear Transformation</title>

<p>For a linear transformation $\ltdefn{T}{U}{V}$, the range is a subset of the codomain $V$.  Informally, it is the set of all outputs that the transformation creates when fed every possible input from the domain.  It will have some natural connections with the column space of a matrix, so we will keep the same notation, and if you think about your objects, then there should be little confusion.  Here's the careful definition.</p>

<definition acro="RLT" index="range!of a linear transformation">
<title>Range of a Linear Transformation</title>
<p>Suppose $\ltdefn{T}{U}{V}$ is a linear transformation.  Then the <define>range</define> of $T$ is the set
<equation>
\rng{T}=\setparts{\lt{T}{\vect{u}}}{\vect{u}\in U}
</equation>
</p>

<notation acro="RLT" index="range">
<title>Range of a Linear Transformation</title>
<usage>$\rng{T}$</usage>
</notation>
</definition>

<example acro="RAO" index="range!linear transformation">
<title>Range, Archetype O</title>

<p><acroref type="archetype" acro="O" /> is the linear transformation
<equation>
<archetypepart acro="O" part="ltdefn" /></equation>
</p>

<p>To determine the elements of $\complex{5}$ in $\rng{T}$, find those vectors $\vect{v}$ such that $\lt{T}{\vect{u}}=\vect{v}$ for some $\vect{u}\in\complex{3}$,
<alignmath>
<![CDATA[\vect{v}&=\lt{T}{\vect{u}}\\]]>
<![CDATA[&=\colvector{-u_1 + u_2 - 3 u_3\\]]>
-u_1 + 2 u_2 - 4 u_3\\
u_1 + u_2 + u_3\\
2 u_1 + 3 u_2 + u_3\\
u_1 + 2 u_3
}\\
<![CDATA[&=]]>
\colvector{-u_1\\-u_1\\u_1\\2 u_1\\ u_1}
+
\colvector{u_2\\2u_2\\u_2\\3u_2\\ 0}
+
\colvector{-3u_3\\-4u_3\\u_3\\u_3\\ 2 u_3}\\
<![CDATA[&=]]>
u_1\colvector{-1\\-1\\1\\2\\1}
+
u_2\colvector{1\\2\\1\\3\\ 0}
+
u_3\colvector{-3\\-4\\1\\1\\2}
</alignmath>
</p>

<p>This says that every output of $T$ ($\vect{v}$) can be written as a linear combination of the three vectors
<alignmath>
\colvector{-1\\-1\\1\\2\\1}
<![CDATA[&&]]>
\colvector{1\\2\\1\\3\\ 0}
<![CDATA[&&]]>
\colvector{-3\\-4\\1\\1\\2}
</alignmath>
using the scalars $u_1,\,u_2,\,u_3$.  Furthermore, since $\vect{u}$ can be any element of $\complex{3}$, every such linear combination is an output.  This means that
<equation>
\rng{T}=\spn{\set{
\colvector{-1\\-1\\1\\2\\1},\,
\colvector{1\\2\\1\\3\\ 0},\,
\colvector{-3\\-4\\1\\1\\2}
}}
</equation>
</p>

<p>The three vectors in this spanning set for $\rng{T}$ form a linearly dependent set (check this!).  So we can find a more economical presentation by any of the various methods from <acroref type="section" acro="CRS" /> and <acroref type="section" acro="FS" />.  We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to <acroref type="theorem" acro="BRS" />, so we can describe the range of $T$ with a basis,
<equation>
\rng{T}=\spn{<archetypepart acro="O" part="ltrangebasis" />}</equation>
</p>

</example>

<p>We know that the span of a set of vectors is always a subspace (<acroref type="theorem" acro="SSS" />), so the range computed in <acroref type="example" acro="RAO" /> is also a subspace.  This is no accident, the range of a linear transformation is <em>always</em> a subspace.</p>

<theorem acro="RLTS" index="range!subspace">
<title>Range of a Linear Transformation is a Subspace</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation.  Then the range of $T$, $\rng{T}$, is a subspace of $V$.</p>

</statement>

<proof>
<p>We can apply the three-part test of <acroref type="theorem" acro="TSS" />.  First, $\zerovector_U\in U$ and $\lt{T}{\zerovector_U}=\zerovector_V$ by <acroref type="theorem" acro="LTTZZ" />, so $\zerovector_V\in\rng{T}$ and we know that the range is non-empty.</p>

<p>Suppose we assume that $\vect{x},\,\vect{y}\in\rng{T}$.  Is $\vect{x}+\vect{y}\in\rng{T}$?  If $\vect{x},\,\vect{y}\in\rng{T}$ then we know there are vectors $\vect{w},\,\vect{z}\in U$ such that $\lt{T}{\vect{w}}=\vect{x}$ and $\lt{T}{\vect{z}}=\vect{y}$.  Because $U$ is a vector space, additive closure (<acroref type="property" acro="AC" />) implies that $\vect{w}+\vect{z}\in U$.</p>

<p>Then
<alignmath>
<![CDATA[\lt{T}{\vect{w}+\vect{z}}&=\lt{T}{\vect{w}}+\lt{T}{\vect{z}}&&]]>\text{<acroref type="definition" acro="LT" />}\\
<![CDATA[&=\vect{x}+\vect{y}&&\text{Definition of $\vect{w}$ and $\vect{z}$}]]>
</alignmath>
</p>

<p>So we have found an input, $\vect{w}+\vect{z}$, which when fed into $T$ creates $\vect{x}+\vect{y}$ as an output.  This qualifies $\vect{x}+\vect{y}$ for membership in $\rng{T}$.  So we have additive closure.</p>

<p>Suppose we assume that $\alpha\in\complex{\null}$ and $\vect{x}\in\rng{T}$.  Is $\alpha\vect{x}\in\rng{T}$?  If $\vect{x}\in\rng{T}$, then there is a vector $\vect{w}\in U$ such that $\lt{T}{\vect{w}}=\vect{x}$.  Because $U$ is a vector space, scalar closure implies that $\alpha\vect{w}\in U$.  Then
<alignmath>
<![CDATA[\lt{T}{\alpha\vect{w}}&=\alpha\lt{T}{\vect{w}}&&]]>\text{<acroref type="definition" acro="LT" />}\\
<![CDATA[&=\alpha\vect{x}&&\text{Definition of $\vect{w}$}]]>
</alignmath>
</p>

<p>So we have found an input ($\alpha\vect{w}$) which when fed into $T$ creates $\alpha\vect{x}$ as an output.  This qualifies $\alpha\vect{x}$ for membership in $\rng{T}$.  So we have scalar closure and <acroref type="theorem" acro="TSS" /> tells us that $\rng{T}$ is a subspace of $V$.</p>

</proof>
</theorem>

<p>Let's compute another range, now that we know in advance that it will be a subspace.</p>

<example acro="FRAN" index="range!full">
<title>Full range, Archetype N</title>

<p><acroref type="archetype" acro="N" /> is the linear transformation
<equation>
<archetypepart acro="N" part="ltdefn" /></equation>
</p>

<p>To determine the elements of $\complex{3}$ in $\rng{T}$, find those vectors $\vect{v}$ such that $\lt{T}{\vect{u}}=\vect{v}$ for some $\vect{u}\in\complex{5}$,
<alignmath>
<![CDATA[\vect{v}&=\lt{T}{\vect{u}}\\]]>
<![CDATA[&=]]>
\colvector{
2 u_1 + u_2 + 3 u_3 - 4 u_4 + 5 u_5\\
u_1 - 2 u_2 + 3 u_3 - 9 u_4 + 3 u_5\\
3 u_1 + 4 u_3 - 6 u_4 + 5 u_5}\\
<![CDATA[&=]]>
\colvector{2u_1\\u_1\\3u_1}+
\colvector{u_2\\-2u_2\\0}+
\colvector{3u_3\\3u_3\\4u_3}+
\colvector{-4u_4\\-9u_4\\-6u_4}+
\colvector{5u_5\\3u_5\\5u_5}\\
<![CDATA[&=]]>
u_1\colvector{2\\1\\3}+
u_2\colvector{1\\-2\\0}+
u_3\colvector{3\\3\\4}+
u_4\colvector{-4\\-9\\-6}+
u_5\colvector{5\\3\\5}\\
</alignmath>
</p>

<p>This says that every output of $T$ ($\vect{v}$) can be written as a linear combination of the five vectors
<alignmath>
<![CDATA[\colvector{2\\1\\3}&&]]>
<![CDATA[\colvector{1\\-2\\0}&&]]>
<![CDATA[\colvector{3\\3\\4}&&]]>
<![CDATA[\colvector{-4\\-9\\-6}&&]]>
\colvector{5\\3\\5}
</alignmath>
using the scalars $u_1,\,u_2,\,u_3,\,u_4,\,u_5$.  Furthermore, since $\vect{u}$ can be any element of $\complex{5}$, every such linear combination is an output.  This means that
<equation>
\rng{T}=\spn{\set{
\colvector{2\\1\\3},\,
\colvector{1\\-2\\0},\,
\colvector{3\\3\\4},\,
\colvector{-4\\-9\\-6},\,
\colvector{5\\3\\5}
}}
</equation>
</p>

<p>The five vectors in this spanning set for $\rng{T}$ form a linearly dependent set (<acroref type="theorem" acro="MVSLD" />).  So we can find a more economical presentation by any of the various methods from <acroref type="section" acro="CRS" /> and <acroref type="section" acro="FS" />.  We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to <acroref type="theorem" acro="BRS" />, so we can describe the range of $T$ with a (nice) basis,
<equation>
\rng{T}=\spn{<archetypepart acro="N" part="ltrangebasis" />}=\complex{3}</equation>
</p>

</example>

<p>In contrast to injective linear transformations having small (trivial) kernels (<acroref type="theorem" acro="KILT" />), surjective linear transformations have large ranges, as indicated in the next theorem.</p>

<theorem acro="RSLT" index="range!surjective linear transformation">
<title>Range of a Surjective Linear Transformation</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation.  Then $T$ is surjective if and only if the range of $T$ equals the codomain, $\rng{T}=V$.</p>

</statement>

<proof>
<p><implyforward /> By <acroref type="definition" acro="RLT" />, we know that $\rng{T}\subseteq V$.  To establish the reverse inclusion, assume $\vect{v}\in V$.  Then since $T$ is surjective (<acroref type="definition" acro="SLT" />), there exists a vector $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$.  However, the existence of $\vect{u}$ gains $\vect{v}$ membership in $\rng{T}$, so $V\subseteq\rng{T}$.  Thus, $\rng{T}=V$.</p>

<p><implyreverse />  To establish that $T$ is surjective, choose $\vect{v}\in V$.  Since we are assuming that $\rng{T}=V$, $\vect{v}\in\rng{T}$.  This says there is a vector $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$, <ie /> $T$ is surjective.</p>

</proof>
</theorem>

<example acro="NSAQR" index="surjective!not">
<title>Not surjective, Archetype Q, revisited</title>

<p>We are now in a position to revisit our first example in this section, <acroref type="example" acro="NSAQ" />.  In that example, we showed that <acroref type="archetype" acro="Q" /> is not surjective by constructing a vector in the codomain where no element of the domain could be used to evaluate the linear transformation to create the output, thus violating <acroref type="definition" acro="SLT" />.  Just where did this vector come from?</p>

<p>The short answer is that the vector
<equation>
\vect{v}=\colvector{-1\\2\\3\\-1\\4}
</equation>
was constructed to lie outside of the range of $T$.  How was this accomplished?  First, the range of $T$ is given by
<equation>
\rng{T}=\spn{<archetypepart acro="Q" part="ltrangebasis" />}</equation>
</p>

<p>Suppose an element of the range $\vect{v^*}$ has its first 4 components equal to $-1$, $2$, $3$, $-1$, in that order.  Then to be an element of $\rng{T}$, we would have
<equation>
\vect{v^*}=(-1)\colvector{1\\0\\0\\0\\1}+(2)\colvector{0\\1\\0\\0\\-1}+(3)
\colvector{0\\0\\1\\0\\-1}+(-1)\colvector{0\\0\\0\\1\\2}
=\colvector{-1\\2\\3\\-1\\-8}
</equation>
</p>

<p>So the only vector in the range with these first four components specified, must have $-8$ in the fifth component.  To set the fifth component to any other value (say, 4) will result in a vector ($\vect{v}$ in <acroref type="example" acro="NSAQ" />)  outside of the range.  Any attempt to find an input for $T$ that will produce $\vect{v}$ as an output will be doomed to failure.</p>

<p>Whenever the range of a linear transformation is not the whole codomain, we can employ this device and conclude that the linear transformation is not surjective.
This is another way of viewing <acroref type="theorem" acro="RSLT" />.  For a surjective linear transformation, the range is all of the codomain and there is no choice for a vector $\vect{v}$ that lies in $V$, yet not in the range.  For every one of the archetypes that is not surjective, there is an example presented of exactly this form.</p>

</example>

<example acro="NSAO" index="surjective!not, Archetype O">
<title>Not surjective, Archetype O</title>

<p>In <acroref type="example" acro="RAO" /> the range of <acroref type="archetype" acro="O" /> was determined to be
<equation>
\rng{T}=\spn{<archetypepart acro="O" part="ltrangebasis" />}</equation>
a subspace of dimension 2 in $\complex{5}$.  Since $\rng{T}\neq\complex{5}$, <acroref type="theorem" acro="RSLT" /> says $T$ is not surjective.</p>

</example>

<example acro="SAN" index="surjective!Archetype N">
<title>Surjective, Archetype N</title>

<p>The range of <acroref type="archetype" acro="N" /> was computed in <acroref type="example" acro="FRAN" /> to be
<equation>
\rng{T}=\spn{<archetypepart acro="N" part="ltrangebasis" />}</equation>
</p>

<p>Since the basis for this subspace is the set of standard unit vectors for $\complex{3}$ (<acroref type="theorem" acro="SUVB" />), we have $\rng{T}=\complex{3}$ and by <acroref type="theorem" acro="RSLT" />, $T$ is  surjective.</p>

</example>

<sageadvice acro="SLT" index="linear transformation!surjective">
<title>Surjective Linear Transformations</title>
The situation in Sage for surjective linear transformations is similar to that for injective linear transformations.  One distinction <mdash /> what your text calls the range of a linear transformation is called the image of a transformation, obtained with the <code>.image()</code> method.  Sage's term is more commonly used, so you are likey to see it again.  As before, two examples, first up is <acroref type="example" acro="RAO" />.
<sage>
<input>U = QQ^3
V = QQ^5
x1, x2, x3 = var('x1, x2, x3')
outputs = [ -x1 +   x2 - 3*x3,
            -x1 + 2*x2 - 4*x3,
             x1 +   x2 +   x3,
           2*x1 + 3*x2 +   x3,
             x1        + 2*x3]
T_symbolic(x1, x2, x3) = outputs
T = linear_transformation(U, V, T_symbolic)
T.is_surjective()
</input>
<output>False
</output>
</sage>

<sage>
<input>T.image()
</input>
<output>Vector space of degree 5 and dimension 2 over Rational Field
Basis matrix:
[ 1  0 -3 -7 -2]
[ 0  1  2  5  1]
</output>
</sage>

Besides showing the relevant commands in action, this demonstrates one half of <acroref type="theorem" acro="RSLT" />.  Now a reprise of <acroref type="example" acro="FRAN" />.
<sage>
<input>U = QQ^5
V = QQ^3
x1, x2, x3, x4, x5 = var('x1, x2, x3, x4, x5')
outputs = [2*x1 +   x2 + 3*x3 - 4*x4 + 5*x5,
             x1 - 2*x2 + 3*x3 - 9*x4 + 3*x5,
           3*x1        + 4*x3 - 6*x4 + 5*x5]
S_symbolic(x1, x2, x3, x4, x5) = outputs
S = linear_transformation(U, V, S_symbolic)
S.is_surjective()
</input>
<output>True
</output>
</sage>

<sage>
<input>S.image()
</input>
<output>Vector space of degree 3 and dimension 3 over Rational Field
Basis matrix:
[1 0 0]
[0 1 0]
[0 0 1]
</output>
</sage>

<sage>
<input>S.image() == V
</input>
<output>True
</output>
</sage>

Previously, we have chosen elements of the codomain which have non-empty or empty preimages.  We can now explain how to do this predictably.  <acroref type="theorem" acro="RPI" /> explains that elements of the codomain with non-empty pre-images are precisely elements of the image.  Consider the non-surjective linear transformation <code>T</code> from above.
<sage>
<input>TI = T.image()
B = TI.basis()
B
</input>
<output>[
(1, 0, -3, -7, -2),
(0, 1, 2, 5, 1)
]
</output>
</sage>

<sage>
<input>b0 = B[0]
b1 = B[1]
</input>
</sage>

Now any linear combination of the basis vectors <code>b0</code> and <code>b1</code> must be an element of the image.  Moreover, the first two slots of the resulting vector will equal the two scalars we choose to create the linear combination.  But most importantly, see that the three remaining slots will be uniquely determined by these two choices.  This means there is exactly one vector in the image with these values in the first two slots.  So if we construct a new vector with these two values in the first two slots, and make any part of the last three slots even slightly different, we will form a vector that cannot be in the image, and will thus have a preimage that is an empty set.  Whew, that is probably worth reading carefully several times, perhaps in conjunction with the example following.
<sage>
<input>in_image = 4*b0 + (-5)*b1
in_image
</input>
<output>(4, -5, -22, -53, -13)
</output>
</sage>

<sage>
<input>T.preimage_representative(in_image)
</input>
<output>(-13, -9, 0)
</output>
</sage>

<sage>
<input>outside_image = 4*b0 + (-5)*b1 + vector(QQ, [0, 0, 0, 0, 1])
outside_image
</input>
<output>(4, -5, -22, -53, -12)
</output>
</sage>

<sage>
<input>T.preimage_representative(outside_image)
</input>
<output>Traceback (most recent call last):
...
ValueError: element is not in the image
</output>
</sage>



</sageadvice>
</subsection>

<subsection acro="SSSLT">
<title>Spanning Sets and Surjective Linear Transformations</title>

<p>Just as injective linear transformations are allied with linear independence (<acroref type="theorem" acro="ILTLI" />, <acroref type="theorem" acro="ILTB" />), surjective linear transformations are allied with spanning sets.</p>

<theorem acro="SSRLT" index="linear transformation!spanning range">
<title>Spanning Set for Range of a Linear Transformation</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and
<alignmath>
<![CDATA[S&=\set{\vectorlist{u}{t}}]]>
</alignmath>
spans $U$.  Then
<alignmath>
<![CDATA[R&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}}]]>
</alignmath>
spans $\rng{T}$.</p>

</statement>

<proof>
<p>We need to establish that $\rng{T}=\spn{R}$, a set equality.  First we establish that $\rng{T}\subseteq\spn{R}$.  To this end, choose $\vect{v}\in\rng{T}$.  Then there exists a vector $\vect{u}\in U$, such that $\lt{T}{\vect{u}}=\vect{v}$ (<acroref type="definition" acro="RLT" />).  Because $S$ spans $U$ there are scalars, $\scalarlist{a}{t}$, such that
<equation>
\vect{u}=\lincombo{a}{u}{t}
</equation>
</p>

<p>Then
<alignmath>
\vect{v}
<![CDATA[&=\lt{T}{\vect{u}}&&]]>\text{<acroref type="definition" acro="RLT" />}\\
<![CDATA[&=\lt{T}{\lincombo{a}{u}{t}}&&]]>\text{<acroref type="definition" acro="SSVS" />}\\
<![CDATA[&=a_1\lt{T}{\vect{u}_1}+a_2\lt{T}{\vect{u}_2}+a_3\lt{T}{\vect{u}_3}+\ldots+a_t\lt{T}{\vect{u}_t}&&]]>\text{<acroref type="theorem" acro="LTLC" />}\\
</alignmath>
which establishes that $\vect{v}\in\spn{R}$ (<acroref type="definition" acro="SS" />).  So $\rng{T}\subseteq\spn{R}$.</p>

<p>To establish the opposite inclusion, choose an element of the span of $R$, say $\vect{v}\in\spn{R}$.  Then there are scalars $\scalarlist{b}{t}$ so that
<alignmath>
\vect{v}
<![CDATA[&=b_1\lt{T}{\vect{u}_1}+b_2\lt{T}{\vect{u}_2}+b_3\lt{T}{\vect{u}_3}+\cdots+b_t\lt{T}{\vect{u}_t}]]>
<![CDATA[&&]]>\text{<acroref type="definition" acro="SS" />}\\
<![CDATA[&=\lt{T}{\lincombo{b}{\vect{u}}{t}}&&]]>\text{<acroref type="theorem" acro="LTLC" />}
</alignmath>
</p>

<p>This demonstrates that $\vect{v}$ is an output of the linear transformation $T$, so $\vect{v}\in\rng{T}$.  Therefore $\spn{R}\subseteq\rng{T}$, so we have the set equality $\rng{T}=\spn{R}$ (<acroref type="definition" acro="SE" />).  In other words, $R$ spans $\rng{T}$ (<acroref type="definition" acro="SSVS" />).</p>

</proof>
</theorem>

<p><acroref type="theorem" acro="SSRLT" /> provides an easy way to begin the construction of a basis for the range of a linear transformation, since the construction of a spanning set requires simply evaluating the linear transformation on a spanning set of the domain.  In practice the best choice for a spanning set of the domain would be as small as possible, in other words, a basis.  The resulting spanning set for the codomain may not be linearly independent, so to find a basis for the range might require tossing out redundant vectors from the spanning set.  Here's an example.</p>

<example acro="BRLT" index="linear transformation!basis of range">
<title>A basis for the range of a linear transformation</title>

<p>Define the linear transformation $\ltdefn{T}{M_{22}}{P_2}$ by
<equation>
<![CDATA[\lt{T}{ \begin{bmatrix} a&b\\c&d \end{bmatrix}}]]>
=\left(a+2b+8c+d\right)+\left(-3a+2b+5d\right)x+\left(a+b+5c\right)x^2
</equation>
</p>

<p>A convenient spanning set for $M_{22}$ is the basis
<equation>
S=\set{
<![CDATA[\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\,]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}]]>
}
</equation>
</p>

<p>So by <acroref type="theorem" acro="SSRLT" />, a spanning set for $\rng{T}$ is
<alignmath>
R
<![CDATA[&=\set{]]>
<![CDATA[\lt{T}{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}},\,]]>
<![CDATA[\lt{T}{\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}},\,]]>
<![CDATA[\lt{T}{\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}},\,]]>
<![CDATA[\lt{T}{\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}}]]>
}\\
<![CDATA[&=\set{1-3x+x^2,\,2+2x+x^2,\,8+5x^2,\,1+5x}]]>
</alignmath>
</p>

<p>The set $R$ is not linearly independent, so if we desire a basis for $\rng{T}$, we need to eliminate some redundant vectors.  Two particular relations of linear dependence on $R$ are
<alignmath>
<![CDATA[(-2)(1-3x+x^2)+(-3)(2+2x+x^2)+(8+5x^2)&=0+0x+0x^2=\zerovector\\]]>
<![CDATA[(1-3x+x^2)+(-1)(2+2x+x^2)+(1+5x)&=0+0x+0x^2=\zerovector]]>
</alignmath>
</p>

<p>These, individually, allow us to remove $8+5x^2$ and $1+5x$ from $R$ without destroying the property that $R$ spans $\rng{T}$.  The two remaining vectors are linearly independent (check this!), so we can write
<equation>
\rng{T}=\spn{\set{1-3x+x^2,\,2+2x+x^2}}
</equation>
and see that $\dimension{\rng{T}}=2$.</p>

</example>

<p>Elements of the range are precisely those elements of the codomain with non-empty preimages.</p>

<theorem acro="RPI" index="range!pre-image">
<title>Range and Pre-Image</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation.  Then
<equation>
\vect{v}\in\rng{T}\text{ if and only if }\preimage{T}{\vect{v}}\neq\emptyset
</equation>
</p>
</statement>

<proof>
<p><implyforward />  If $\vect{v}\in\rng{T}$, then there is a vector $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$.  This qualifies $\vect{u}$ for membership in $\preimage{T}{\vect{v}}$, and thus the preimage of $\vect{v}$ is not empty.</p>

<p><implyreverse />  Suppose the preimage of $\vect{v}$ is not empty, so we can choose a vector $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$.  Then $\vect{v}\in\rng{T}$.</p>
</proof>
</theorem>

<p>Now would be a good time to return to <acroref type="diagram" acro="KPI" /> which depicted the pre-images of a non-surjective linear transformation.  The vectors $\vect{x},\,\vect{y}\in V$ were elements of the codomain whose pre-images were empty, as we expect for a non-surjective linear transformation from the characterization in <acroref type="theorem" acro="RPI" />.</p>

<theorem acro="SLTB" index="surjective linear transformation!bases">
<title>Surjective Linear Transformations and Bases</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and
<alignmath>
<![CDATA[B&=\set{\vectorlist{u}{m}}]]>
</alignmath>
is a basis of $U$.  Then $T$ is surjective if and only if
<alignmath>
<![CDATA[C&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}]]>
</alignmath>
is a spanning set for $V$.</p>
</statement>

<proof>
<p><implyforward />  Assume $T$ is surjective.  Since $B$ is a basis, we know $B$ is a spanning set of $U$ (<acroref type="definition" acro="B" />).  Then <acroref type="theorem" acro="SSRLT" /> says that $C$ spans $\rng{T}$.  But the hypothesis that $T$ is surjective means $V=\rng{T}$ (<acroref type="theorem" acro="RSLT" />), so $C$ spans $V$.</p>

<p><implyreverse />  Assume that $C$ spans $V$.  To establish that $T$ is surjective, we will show that every element of $V$ is an output of $T$ for some input (<acroref type="definition" acro="SLT" />).  Suppose that $\vect{v}\in V$.  As an element of $V$, we can write $\vect{v}$ as a linear combination of the spanning set $C$.  So there are are scalars, $\scalarlist{b}{m}$, such that
<equation>
\vect{v}=b_1\lt{T}{\vect{u}_1}+b_2\lt{T}{\vect{u}_2}+b_3\lt{T}{\vect{u}_3}+\cdots+b_m\lt{T}{\vect{u}_m}
</equation>
</p>

<p>Now define the vector $\vect{u}\in U$ by
<equation>
\vect{u}=\lincombo{b}{u}{m}
</equation>
</p>

<p>Then
<alignmath>
<![CDATA[\lt{T}{\vect{u}}&=\lt{T}{\lincombo{b}{u}{m}}\\]]>
<![CDATA[&=b_1\lt{T}{\vect{u}_1}+b_2\lt{T}{\vect{u}_2}+b_3\lt{T}{\vect{u}_3}+\cdots+b_m\lt{T}{\vect{u}_m}&&]]>\text{<acroref type="theorem" acro="LTLC" />}\\
<![CDATA[&=\vect{v}]]>
</alignmath>
</p>

<p>So, given any choice of a vector $\vect{v}\in V$, we can design an input $\vect{u}\in U$ to produce $\vect{v}$ as an output of $T$.  Thus, by <acroref type="definition" acro="SLT" />, $T$ is surjective.</p>

</proof>
</theorem>

</subsection>

<subsection acro="SLTD">
<title>Surjective Linear Transformations and Dimension</title>

<theorem acro="SLTD" index="surjective linear transformations!dimension">
<title>Surjective Linear Transformations and Dimension</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ is a surjective linear transformation.  Then $\dimension{U}\geq\dimension{V}$.</p>

</statement>

<proof>
<p>Suppose to the contrary that
<![CDATA[$m=\dimension{U}<\dimension{V}=t$.]]>
Let $B$ be  a basis of $U$, which will then contain $m$ vectors.  Apply $T$ to each element of $B$ to form a set $C$ that is a subset of $V$.  By <acroref type="theorem" acro="SLTB" />, $C$ is a spanning set of $V$ with $m$ or fewer vectors.  So we have a set of $m$ or fewer vectors that span $V$, a vector space of dimension $t$, with
<![CDATA[$m<t$.]]>
However, this contradicts <acroref type="theorem" acro="G" />, so our assumption is false and $\dimension{U}\geq\dimension{V}$.</p>

</proof>
</theorem>

<example acro="NSDAT" index="surjective!not, by dimension">
<title>Not surjective by dimension, Archetype T</title>

<p>The linear transformation in <acroref type="archetype" acro="T" /> is
<equation>
<archetypepart acro="T" part="ltdefn" /></equation></p>

<p>Since
<![CDATA[$\dimension{P_4}=5<6=\dimension{P_5}$,]]>
$T$ cannot be surjective for then it would violate <acroref type="theorem" acro="SLTD" />.</p>

</example>

<p>Notice that the previous example made no use of the actual formula defining the function.  Merely a comparison of the dimensions of the domain and codomain are enough to conclude that the linear transformation is not surjective.  <acroref type="archetype" acro="O" /> and <acroref type="archetype" acro="P" /> are two more examples of linear transformations that have <q>small</q> domains and <q>big</q> codomains, resulting in an inability to create all possible outputs and thus they are non-surjective linear transformations.</p>

</subsection>

<subsection acro="CSLT">
<title>Composition of Surjective Linear Transformations</title>

<p>In <acroref type="subsection" acro="LT.NLTFO" /> we saw how to combine linear transformations to build new linear transformations, specifically, how to build the composition of two linear transformations (<acroref type="definition" acro="LTC" />).  It will be useful later to know that the composition of surjective linear transformations is again surjective, so we prove that here.</p>

<theorem acro="CSLTS" index="composition!surjective linear transformations">
<title>Composition of Surjective Linear Transformations is Surjective</title>
<statement>
<p>Suppose that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are surjective linear transformations.  Then $\ltdefn{(\compose{S}{T})}{U}{W}$ is a surjective linear transformation.</p>

</statement>

<proof>
<p>That the composition is a linear transformation was established in <acroref type="theorem" acro="CLTLT" />, so we need only establish that the composition is surjective.  Applying <acroref type="definition" acro="SLT" />, choose $\vect{w}\in W$.</p>

<p>Because $S$ is surjective, there must be a vector $\vect{v}\in V$, such that $\lt{S}{\vect{v}}=\vect{w}$.  With the existence of $\vect{v}$ established, that $T$ is surjective guarantees a vector $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$.  Now,
<alignmath>
<![CDATA[\lt{\left(\compose{S}{T}\right)}{\vect{u}}&=\lt{S}{\lt{T}{\vect{u}}}&&]]>\text{<acroref type="definition" acro="LTC" />}\\
<![CDATA[&=\lt{S}{\vect{v}}&&\text{Definition of $\vect{u}$}\\]]>
<![CDATA[&=\vect{w}&&\text{Definition of $\vect{v}$}]]>
</alignmath>
</p>

<p>This establishes that any element of the codomain ($\vect{w}$) can be created by evaluating $\compose{S}{T}$ with the right input ($\vect{u}$).  Thus, by <acroref type="definition" acro="SLT" />, $\compose{S}{T}$ is surjective.</p>

</proof>
</theorem>

<sageadvice acro="CSLT" index="linear transformation!composition">
<title>Composition of Surjective Linear Transformations</title>
As we mentioned in the last section, experimenting with Sage is a worthwhile complement to other methods of learning mathematics.  We have purposely avoided providing illustrations of deeper results, such as <acroref type="theorem" acro="ILTB" /> and  <acroref type="theorem" acro="SLTB" />, which you should now be equipped to investigate yourself.  For completeness, and since composition wil be very important in the next few sections, we will provide an illustration of <acroref type="theorem" acro="CSLTS" />.  Similar to what we did in the previous section, we choose dimensions suggested by <acroref type="theorem" acro="SLTD" />, and then use randomly constructed matrices to form a pair of surjective linear transformations.
<sage>
<input>U = QQ^4
V = QQ^3
W = QQ^2
A = matrix(QQ, 3, 4, [[ 3, -2,  8, -9],
                      [-1,  3, -4, -1],
                      [ 3,  2,  8,  3]])
T = linear_transformation(U, V, A, side='right')
T.is_surjective()
</input>
<output>True
</output>
</sage>

<sage>
<input>B = matrix(QQ, 2, 3, [[ 8, -5, 3],
                      [-2,  1, 1]])
S = linear_transformation(V, W, B, side='right')
S.is_surjective()
</input>
<output>True
</output>
</sage>

<sage>
<input>C = S*T
C.is_surjective()
</input>
<output>True
</output>
</sage>



</sageadvice>
</subsection>

<!--   End of  slt.tex -->
<readingquestions>
<ol>
<li>Suppose $\ltdefn{T}{\complex{5}}{\complex{8}}$ is a linear transformation.  Why can't $T$ be surjective?
</li>
<li>What is the relationship between a surjective linear transformation and its range?
</li>
<li>There are many similarities and differences between injective and surjective linear transformations.  Compare and contrast these two different types of linear transformations.  (This means going well beyond just stating their definitions.)
</li></ol>
</readingquestions>

<exercisesubsection>

<exercise type="C" number="10" rough="range for each archetype">
<problem contributor="robertbeezer">Each archetype below is a linear transformation.  Compute the range for each.<br /><br />
<acroref type="archetype" acro="M" />,
<acroref type="archetype" acro="N" />,
<acroref type="archetype" acro="O" />,
<acroref type="archetype" acro="P" />,
<acroref type="archetype" acro="Q" />,
<acroref type="archetype" acro="R" />,
<acroref type="archetype" acro="S" />,
<acroref type="archetype" acro="T" />,
<acroref type="archetype" acro="U" />,
<acroref type="archetype" acro="V" />,
<acroref type="archetype" acro="W" />,
<acroref type="archetype" acro="X" />
<!-- % TODO:  Check competeness when Y,Z appear. -->
</problem>
</exercise>

<exercise type="C" number="20" rough="verify surjectivity in Example SAR">
<problem contributor="robertbeezer"><acroref type="example" acro="SAR" /> concludes with an expression for a vector $\vect{u}\in\complex{5}$ that we believe will create the vector $\vect{v}\in\complex{5}$ when used to evaluate $T$.  That is, $\lt{T}{\vect{u}}=\vect{v}$.  Verify this assertion by actually evaluating $T$ with $\vect{u}$.  If you don't have the patience to push around all these symbols, try choosing a numerical instance of $\vect{v}$, compute $\vect{u}$, and then compute $\lt{T}{\vect{u}}$, which should result in $\vect{v}$.
</problem>
</exercise>

<exercise type="C" number="22" rough="Non-surjective, find empty pre-image">
<problem contributor="robertbeezer">The linear transformation $\ltdefn{S}{\complex{4}}{\complex{3}}$ is not surjective.  Find an output $\vect{w}\in\complex{3}$ that has an empty pre-image (that is $\preimage{S}{\vect{w}}=\emptyset$.)
<equation>
\lt{S}{\colvector{x_1\\x_2\\x_3\\x_4}}=
\colvector{
2x_1+x_2+3x_3-4x_4\\
x_1+3x_2+4x_3+3x_4\\
-x_1+2x_2+x_3+7x_4
}
</equation>
</problem>
<solution contributor="robertbeezer">To find an element of $\complex{3}$ with an empty pre-image, we will compute the range of the linear transformation $\rng{S}$ and then find an element outside of this set.<br /><br />
By <acroref type="theorem" acro="SSRLT" /> we can evaluate $S$ with the elements of a spanning set of the domain and create a spanning set for the range.
<alignmath>
<![CDATA[\lt{S}{\colvector{1\\0\\0\\0}}&=\colvector{2\\1\\-1}]]>
<![CDATA[&]]>
<![CDATA[\lt{S}{\colvector{0\\1\\0\\0}}&=\colvector{1\\3\\2}]]>
<![CDATA[&]]>
<![CDATA[\lt{S}{\colvector{0\\0\\1\\0}}&=\colvector{3\\4\\1}]]>
<![CDATA[&]]>
<![CDATA[\lt{S}{\colvector{0\\0\\0\\1}}&=\colvector{-4\\3\\7}]]>
</alignmath>
So
<equation>
\rng{S}=\spn{\set{
\colvector{2\\1\\-1},\,
\colvector{1\\3\\2},\,
\colvector{3\\4\\1},\,
\colvector{-4\\3\\7}
}}
</equation>
This spanning set is obviously linearly dependent, so we can reduce it to a basis for $\rng{S}$ using <acroref type="theorem" acro="BRS" />, where the elements of the spanning set are placed as the rows of a matrix.  The result is that
<equation>
\rng{S}=\spn{\set{
\colvector{1\\0\\-1},\,
\colvector{0\\1\\1}
}}
</equation>
Therefore, the unique vector in $\rng{S}$ with a first slot equal to 6 and a second slot equal to 15 will be the linear combination
<equation>
6\colvector{1\\0\\-1}+15\colvector{0\\1\\1}=\colvector{6\\15\\9}
</equation>
So, any vector with first two components equal to 6 and 15, but with a third component different from 9, such as
<equation>
\vect{w}=\colvector{6\\15\\-63}
</equation>
will not be an element of the range of $S$ and will therefore have an empty pre-image.
Another strategy on this problem is to <em>guess</em>.  Almost any vector will lie outside the range of $T$, you have to be unlucky to randomly choose an element of the range.  This is because the codomain has dimension 3, while the range is <q>much smaller</q> at a dimension of 2.  You still need to check that your guess lies outside of the range, which generally will involve solving a system of equations that turns out to be inconsistent.
</solution>
</exercise>

<exercise type="C" number="23" rough="Is T surjective?">
<problem contributor="chrisblack">Determine whether or not the following linear transformation $\ltdefn{T}{\complex{5}}{P_3}$ is surjective:
<alignmath>
<![CDATA[\lt{T}{\colvector{a\\b\\c\\d\\e}} &= a + (b + c)x + (c + d)x^2 + (d + e)x^3]]>
</alignmath>
</problem>
<solution contributor="chrisblack">The linear transformation $T$ is surjective if for any $p(x) = \alpha + \beta x + \gamma x^2 + \delta x^3$, there is a vector $\vect{u} = \colvector{a\\b\\c\\d\\e}$ in $\complex{5}$ so that $\lt{T}{\vect{u}} = p(x)$.  We need to be able to solve the system
<alignmath>
<![CDATA[a &=\alpha\\]]>
<![CDATA[b+c &= \beta\\]]>
<![CDATA[c + d &= \gamma\\]]>
<![CDATA[d + e &= \delta]]>
</alignmath>
This system has an infinite number of solutions, one of which is $a = \alpha$, $b = \beta$, $c = 0$, $d = \gamma$ and $e = \delta - \gamma$, so that
<alignmath>
\lt{T}{\colvector{\alpha\\ \beta\\0\\ \gamma \\ \delta - \gamma}}
<![CDATA[&= \alpha + (\beta + 0)x + (0 + \gamma) x^2 + (\gamma + (\delta - \gamma)) x^3\\]]>
<![CDATA[&= \alpha + \beta x + \gamma x^2 + \delta x^3\\]]>
<![CDATA[&= p(x).]]>
</alignmath>
Thus, $T$ is surjective, since for every vector $\vect{v} \in P_3$, there exists a vector $\vect{u} \in \complex{5}$ so that $\lt{T}{\vect{u}} = \vect{v}$.
</solution>
</exercise>

<exercise type="C" number="24" rough="Is T surjective?">
<problem contributor="chrisblack">Determine whether or not the linear transformation $\ltdefn{T}{P_3}{\complex{5}}$ below is surjective:
<alignmath>
<![CDATA[\lt{T}{a + bx + cx^2 + dx^3} &= \colvector{a + b \\ b + c \\ c + d \\ a + c\\ b + d}.]]>
</alignmath>
<!-- \[ T(\] -->
</problem>
<solution contributor="chrisblack">According to <acroref type="theorem" acro="SLTD" />, if a linear transformation $\ltdefn{T}{U}{V}$ is surjective, then $\dimension{U}\ge\dimension{V}$.  In this example, $U = P_3$ has dimension 4, and $V = \complex{5}$ has dimension 5, so $T$ cannot be surjective.  (There is no way $T$ can <q>expand</q> the domain $P_3$ to fill the codomain $\complex{5}$.)
</solution>
</exercise>

<exercise type="C" number="25" rough="range for a linear transformation">
<problem contributor="robertbeezer">Define the linear transformation
<equation>
\ltdefn{T}{\complex{3}}{\complex{2}},\quad
\lt{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3}
</equation>
Find a basis for the range of $T$, $\rng{T}$.  Is $T$ surjective?
</problem>
<solution contributor="robertbeezer">To find the range of $T$, apply $T$ to the elements of a spanning set for $\complex{3}$ as suggested in <acroref type="theorem" acro="SSRLT" />.  We will use the standard basis vectors (<acroref type="theorem" acro="SUVB" />).
<equation>
\rng{T}=
\spn{\set{\lt{T}{\vect{e}_1},\,\lt{T}{\vect{e}_2},\,\lt{T}{\vect{e}_3}}}=
\spn{\set{\colvector{2\\-4},\,\colvector{-1\\2},\,\colvector{5\\-10}}}
</equation>
Each of these vectors is a scalar multiple of the others, so we can toss two of them in reducing the spanning set to a linearly independent set (or be more careful and apply <acroref type="theorem" acro="BCS" /> on a matrix with these three vectors as columns).  The result is the basis of the range,
<equation>
\set{\colvector{1\\-2}}
</equation>
<br /><br />
With $\rank{T}\neq 2$, $\rng{T}\neq\complex{2}$, so <acroref type="theorem" acro="RSLT" /> says $T$ is not surjective.
</solution>
</exercise>

<exercise type="C" number="26" rough="Find basis of range.  Is T surjective?">
<problem contributor="chrisblack">Let $\ltdefn{T}{\complex{3}}{\complex{3}} $ be given by
$\lt{T}{\colvector{a\\b\\c}} = \colvector{a + b + 2c\\ 2c\\ a + b + c}$.  Find a basis of $\rng{T}$.  Is $T$ surjective?
</problem>
<solution contributor="chrisblack">The range of $T$ is
<alignmath>
\rng{T}
<![CDATA[&= \setparts{\colvector{a + b + 2c\\ 2c\\a + b + c}}{a,b,c\in\complexes}\\]]>
<![CDATA[&= \setparts{a\colvector{1\\0\\1} + b \colvector{1\\0\\1}+ c\colvector{2\\2\\1}}{a,b,c\in\complexes}\\]]>
<![CDATA[&= \spn{\colvector{1\\0\\1},\colvector{2\\2\\1}}]]>
</alignmath>
Since the vectors $\colvector{1\\0\\1}$ and $\colvector{2\\2\\1}$ are linearly independent (why?), a basis of $\rng{T}$ is $\set{\colvector{1\\0\\1},\colvector{2\\2\\1}}$.  Since the dimension of the range is 2 and the dimension of the codomain is 3, $T$ is not surjective.
</solution>
</exercise>

<exercise type="C" number="27" rough="Find basis of range.  Is T surjective?">
<problem contributor="chrisblack">Let $\ltdefn{T}{\complex{3}}{\complex{4}}$ be given by
$\lt{T}{\colvector{a\\b\\c}} = \colvector{a + b -c\\ a - b + c\\ -a + b + c\\a + b + c}$.
Find a basis of $\rng{T}$.  Is $T$ surjective?
</problem>
<solution contributor="chrisblack">The range of $T$ is
<alignmath>
\rng{T}
<![CDATA[&= \setparts{\colvector{a + b -c \\ a - b + c\\ -a + b + c\\a + b + c}}{a,b,c\in\complexes}\\]]>
<![CDATA[&= \setparts{a\colvector{1\\1\\-1\\1} + b \colvector{1\\-1\\1\\1} + c\colvector{-1\\1\\1\\1}}{a,b,c\in\complexes}\\]]>
<![CDATA[&= \spn{\colvector{1\\1\\-1\\1},\colvector{1\\-1\\1\\1}, \colvector{-1\\1\\1\\1}}]]>
</alignmath>
By row reduction (not shown), we can see that the set
<alignmath>
\set{
\colvector{1\\1\\-1\\1},\,
\colvector{1\\-1\\1\\1},\,
\colvector{-1\\1\\1\\1}
}
</alignmath>
are linearly independent, so is a basis of $\rng{T}$.  Since the dimension of the range is 3 and the dimension of the codomain is 4, $T$ is not surjective.  (We should have anticipated that $T$ was not surjective since the dimension of the domain is smaller than the dimension of the codomain.)
</solution>
</exercise>

<exercise type="C" number="28" rough="Find basis of range.  Is T surjective?">
<problem contributor="chrisblack">Let $\ltdefn{T}{\complex{4}}{ M_{2,2}}$ be given by
<![CDATA[$\lt{T}{\colvector{a\\b\\c\\d}} = \begin{bmatrix} a + b & a + b + c\\ a + b + c & a + d\end{bmatrix}$.]]>
 Find a basis of $\rng{T}$.
Is $T$ surjective?
</problem>
<solution contributor="chrisblack">The range of $T$ is
<alignmath>
\rng{T}
<![CDATA[&= \setparts{]]>
\begin{bmatrix}
<![CDATA[a + b & a + b + c\\]]>
<![CDATA[a + b + c& a + d]]>
\end{bmatrix}}{a,b,c,d\in\complexes}\\
<![CDATA[&= \setparts{]]>
<![CDATA[a\begin{bmatrix} 1 & 1 \\1 & 1 \end{bmatrix} +]]>
<![CDATA[b\begin{bmatrix} 1 & 1 \\1 & 0 \end{bmatrix} +]]>
<![CDATA[c\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix} +]]>
<![CDATA[d\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}]]>
}{a,b,c,d\in\complexes}\\
<![CDATA[&=\spn{]]>
<![CDATA[\begin{bmatrix} 1 & 1 \\1 & 1 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 1 & 1 \\1 & 0 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}}\\]]>
<![CDATA[&=\spn{]]>
<![CDATA[\begin{bmatrix} 1 & 1 \\1 & 0 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}}.]]>
</alignmath>
Can you explain the last equality above?<br /><br />
These three matrices are linearly independent, so a basis of $\rng{T}$ is
$\set{
<![CDATA[\begin{bmatrix} 1 & 1 \\1 & 0 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix},]]>
<![CDATA[\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}}$.  Thus, $T$ is not surjective, since the range has dimension 3 which is shy of $\dimension{M_{2,2}}=4$.]]>
(Notice that the range is actually the subspace of symmetric $2 \times 2$ matrices in $M_{2,2}$.)
</solution>
</exercise>

<exercise type="C" number="29" rough="Find basis of range.  Is T surjective?">
<problem contributor="chrisblack">Let $\ltdefn{T}{P_2}{P_4}$ be given by
$\lt{T}{p(x)} = x^2 p(x)$.
Find a basis of $\rng{T}$.
Is $T$ surjective?
</problem>
<solution contributor="chrisblack">If we transform the basis of $P_2$, then <acroref type="theorem" acro="SSRLT" /> guarantees we will have a spanning set of $\rng{T}$.  A basis of $P_2$ is $\set{1, x, x^2}$.  If we transform the elements of this set, we get the set $\set{x^2, x^3, x^4}$ which is a spanning set for $\rng{T}$.  These three vectors are linearly independent, so $\set{x^2, x^3, x^4}$ is a basis of $\rng{T}$.
</solution>
</exercise>

<exercise type="C" number="30" rough="Find basis of range.  Is T surjective?">
<problem contributor="chrisblack">Let $\ltdefn{T}{P_4}{P_3}$ be given by
$\lt{T}{p(x)} = p^\prime(x)$, where $p^\prime(x)$ is the derivative.
Find a basis of $\rng{T}$.
Is $T$ surjective?
</problem>
<solution contributor="chrisblack">If we transform the basis of $P_4$, then <acroref type="theorem" acro="SSRLT" /> guarantees we will have a spanning set of $\rng{T}$.  A basis of $P_4$ is $\set{1, x, x^2, x^3, x^4}$.  If we transform the elements of this set, we get the set $\set{0, 1, 2x, 3x^2, 4x^3}$ which is a spanning set for $\rng{T}$.  Reducing this to a linearly independent set, we find that $\{1, 2x, 3x^2, 4x^3\}$ is a basis of $\rng{T}$.  Since $\rng{T}$ and $P_3$ both have dimension 4, $T$ is surjective.
</solution>
</exercise>

<exercise type="C" number="40" rough="non-injective by demonstration">
<problem contributor="robertbeezer">Show that the linear transformation $T$ is not surjective by finding an element of the codomain, $\vect{v}$,  such that there is  no vector $\vect{u}$ with $\lt{T}{\vect{u}}=\vect{v}$.
<equation>
\ltdefn{T}{\complex{3}}{\complex{3}},\quad
\lt{T}{\colvector{a\\b\\c}}=
\colvector{2a+3b-c\\2b-2c\\a-b+2c}
</equation>
</problem>
<solution contributor="robertbeezer">We wish to find an output vector $\vect{v}$ that has no associated input.  This is the same as requiring that there is no solution to the equality
<equation>
\vect{v}=\lt{T}{\colvector{a\\b\\c}}=\colvector{2a+3b-c\\2b-2c\\a-b+2c}
=a\colvector{2\\0\\1}+b\colvector{3\\2\\-1}+c\colvector{-1\\-2\\2}
</equation>
In other words, we would like to find an element of $\complex{3}$ not in the set
<equation>
Y=\spn{\set{\colvector{2\\0\\1},\,\colvector{3\\2\\-1},\,\colvector{-1\\-2\\2}}}
</equation>
If we make these vectors the rows of a matrix, and row-reduce, <acroref type="theorem" acro="BRS" /> provides an alternate description of $Y$,
<equation>
Y=\spn{\set{\colvector{2\\0\\1},\,\colvector{0\\4\\-5}}}
</equation>
If we add these vectors together, and then change the third component of the result, we will create a vector that lies outside of $Y$, say $\vect{v}=\colvector{2\\4\\9}$.
</solution>
</exercise>

<exercise type="M" number="60" rough="zero linear transformation">
<problem contributor="robertbeezer">Suppose $U$ and $V$ are vector spaces.  Define the function $\ltdefn{Z}{U}{V}$ by $\lt{T}{\vect{u}}=\zerovector_{V}$ for every $\vect{u}\in U$.  Then by <acroref type="exercise" acro="LT.M60" />, $Z$ is a linear transformation.  Formulate a condition on $V$ that is equivalent to $Z$ being an surjective linear transformation.   In other words, fill in the blank to complete the following statement (and then give a proof):  $Z$ is surjective if and only if $V$ is <rule width="1in" height="1pt" />. (See <acroref type="exercise" acro="ILT.M60" />, <acroref type="exercise" acro="IVLT.M60" />.)
</problem>
</exercise>

<exercise type="T" number="15" rough="composition and range inclusion">
<problem contributor="robertbeezer">Suppose that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are linear transformations.  Prove the following relationship between ranges.
<equation>
\rng{\compose{S}{T}}\subseteq\rng{S}
</equation>
</problem>
<solution contributor="robertbeezer">This question asks us to establish that one set ($\rng{\compose{S}{T}}$) is a subset of another ($\rng{S}$).  Choose an element in the <q>smaller</q> set, say $\vect{w}\in\rng{\compose{S}{T}}$.  Then we know that there is a vector $\vect{u}\in U$ such that
<equation>
\vect{w}=\lt{\left(\compose{S}{T}\right)}{\vect{u}}=\lt{S}{\lt{T}{\vect{u}}}
</equation>
Now define $\vect{v}=\lt{T}{\vect{u}}$, so that then
<equation>
\lt{S}{\vect{v}}=\lt{S}{\lt{T}{\vect{u}}}=\vect{w}
</equation>
This statement is sufficient to show that $\vect{w}\in\rng{S}$, so $\vect{w}$ is an element of the <q>larger</q> set, and $\rng{\compose{S}{T}}\subseteq\rng{S}$.
</solution>
</exercise>

<exercise type="T" number="20" rough="range is column space of matrix defining lt">
<problem contributor="andyzimmer">Suppose that $A$ is an $m\times n$ matrix.  Define the linear transformation $T$ by
<equation>
\ltdefn{T}{\complex{n}}{\complex{m}},\quad \lt{T}{\vect{x}}=A\vect{x}
</equation>
Prove that the range of $T$ equals the column space of $A$, $\rng{T}=\csp{A}$.
</problem>
<solution contributor="andyzimmer">This is an equality of sets, so we want to establish two subset conditions (<acroref type="definition" acro="SE" />).<br /><br />
First, show $\csp{A}\subseteq\rng{T}$.  Choose $\vect{y}\in\csp{A}$.  Then by <acroref type="definition" acro="CSM" /> and <acroref type="definition" acro="MVP" /> there is a vector $\vect{x}\in\complex{n}$ such that $A\vect{x}=\vect{y}$.  Then
<alignmath>
\lt{T}{\vect{x}}
<![CDATA[&=A\vect{x}&&\text{Definition of $T$}\\]]>
<![CDATA[&=\vect{y}]]>
</alignmath>
This statement qualifies $\vect{y}$ as a member of $\rng{T}$ (<acroref type="definition" acro="RLT" />), so $\csp{A}\subseteq\rng{T}$.<br /><br />
Now, show $\rng{T}\subseteq\csp{A}$.  Choose $\vect{y}\in\rng{T}$.  Then by <acroref type="definition" acro="RLT" />, there is a vector $\vect{x}$ in $\complex{n}$ such that $\lt{T}{\vect{x}}=\vect{y}$.  Then
<alignmath>
A\vect{x}
<![CDATA[&=\lt{T}{\vect{x}}&&\text{Definition of $T$}\\]]>
<![CDATA[&=\vect{y}]]>
</alignmath>
So by <acroref type="definition" acro="CSM" /> and <acroref type="definition" acro="MVP" />, $\vect{y}$ qualifies for membership in $\csp{A}$ and so $\rng{T}\subseteq\csp{A}$.<br /><br />
</solution>
</exercise>

</exercisesubsection>

</section>