# HG changeset patch # User Rob Beezer # Date 1352941699 28800 # Node ID eea73425e4fb570e1fcc3e007bf354b393fd6530 # Parent be31a794f0d272f5134c0abc4c76fe040b52e45a Line breaks, Chapter LT diff --git a/src/section-ILT.xml b/src/section-ILT.xml --- a/src/section-ILT.xml +++ b/src/section-ILT.xml @@ -96,42 +96,30 @@

-

To establish that $R$ is injective we must begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$ and somehow arrive from this at the conclusion that $\vect{x}=\vect{y}$. Here we go, - +

To establish that $R$ is injective we must begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$ and somehow arrive at the conclusion that $\vect{x}=\vect{y}$. Here we go, -\lt{T}{\vect{x}}=\lt{T}{\vect{y}}\\ -\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}=\lt{T}{\colvector{y_1\\y_2\\y_3\\y_4\\y_5}}\\ +\colvector{0\\0\\0\\0\\0} +\lt{T}{\vect{x}}-\lt{T}{\vect{y}}\\ +\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}-\lt{T}{\colvector{y_1\\y_2\\y_3\\y_4\\y_5}}\\ + \colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\ 36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\ -44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\ 34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\ -12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5} -= +12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5}\\ + \colvector{-65 y_1 + 128 y_2 + 10 y_3 - 262 y_4 + 40 y_5\\ 36 y_1 - 73 y_2 - y_3 + 151 y_4 - 16 y_5\\ -44 y_1 + 88 y_2 + 5 y_3 - 180 y_4 + 24 y_5\\ 34 y_1 - 68 y_2 - 3 y_3 + 140 y_4 - 18 y_5\\ 12 y_1 - 24 y_2 - y_3 + 49 y_4 - 5 y_5}\\ -\colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\ -36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\ --44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\ -34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\ -12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5} -- -\colvector{-65 y_1 + 128 y_2 + 10 y_3 - 262 y_4 + 40 y_5\\ -36 y_1 - 73 y_2 - y_3 + 151 y_4 - 16 y_5\\ --44 y_1 + 88 y_2 + 5 y_3 - 180 y_4 + 24 y_5\\ -34 y_1 - 68 y_2 - 3 y_3 + 140 y_4 - 18 y_5\\ -12 y_1 - 24 y_2 - y_3 + 49 y_4 - 5 y_5} -= -\colvector{0\\0\\0\\0\\0}\\ + \colvector{-65 (x_1-y_1) + 128 (x_2-y_2) + 10 (x_3-y_3) - 262 (x_4-y_4) + 40 (x_5-y_5)\\ 36 (x_1-y_1) - 73 (x_2-y_2) - (x_3-y_3) + 151 (x_4-y_4) - 16 (x_5-y_5)\\ -44 (x_1-y_1) + 88 (x_2-y_2) + 5 (x_3-y_3) - 180 (x_4-y_4) + 24 (x_5-y_5)\\ 34 (x_1-y_1) - 68 (x_2-y_2) - 3 (x_3-y_3) + 140 (x_4-y_4) - 18 (x_5-y_5)\\ -12 (x_1-y_1) - 24 (x_2-y_2) - (x_3-y_3) + 49 (x_4-y_4) - 5 (x_5-y_5)} -= -\colvector{0\\0\\0\\0\\0}\\ +12 (x_1-y_1) - 24 (x_2-y_2) - (x_3-y_3) + 49 (x_4-y_4) - 5 (x_5-y_5)}\\ + \begin{bmatrix} @@ -140,8 +128,6 @@ \end{bmatrix} \colvector{x_1-y_1\\x_2-y_2\\x_3-y_3\\x_4-y_4\\x_5-y_5} -= -\colvector{0\\0\\0\\0\\0}

@@ -726,8 +712,15 @@ Injective Linear Transformations and Linear Independence -

Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation and $S=\set{\vectorlist{u}{t}}$ is a linearly independent subset of $U$. Then $R=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}}$ is a linearly independent subset of $V$.

- +

Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation and + + + +is a linearly independent subset of $U$. Then + + + +is a linearly independent subset of $V$.

@@ -753,7 +746,15 @@ Injective Linear Transformations and Bases -

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $B=\set{\vectorlist{u}{m}}$ is a basis of $U$. Then $T$ is injective if and only if $C=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}$ is a linearly independent subset of $V$.

+

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and + + + +is a basis of $U$. Then $T$ is injective if and only if + + + +is a linearly independent subset of $V$.

diff --git a/src/section-IVLT.xml b/src/section-IVLT.xml --- a/src/section-IVLT.xml +++ b/src/section-IVLT.xml @@ -60,26 +60,26 @@

Then - - - - + + + + \end{bmatrix}\\ - - + + and - -\lt{S}{\lt{T}{a+bx+cx^2+dx^3}}\\ - + +\\ + \end{bmatrix}}\\ - + (d+(b-d))x\\ (d)x^3\\ - - + +

@@ -884,7 +884,7 @@

-

This says that for any vector, $\vect{u}$, from $U$, there exist scalars ($\scalarlist{d}{s},\,\scalarlist{c}{r}$) that form $\vect{u}$ as a linear combination of the vectors in the set $B$. In other words, $B$ spans $U$ ().

+

This says that for any vector, $\vect{u}$, from $U$, there exist scalars ($\scalarlist{d}{s}$, $\scalarlist{c}{r}$) that form $\vect{u}$ as a linear combination of the vectors in the set $B$. In other words, $B$ spans $U$ ().

So $B$ is a basis () of $U$ with $s+r$ vectors, and thus diff --git a/src/section-LT.xml b/src/section-LT.xml --- a/src/section-LT.xml +++ b/src/section-LT.xml @@ -179,7 +179,8 @@

Then - + +

@@ -214,7 +215,9 @@

Return to and compute $\lt{S}{\colvector{0\\0\\0}}=\colvector{0\\0\\-2}$ to quickly see again that $S$ is not a linear transformation, while in compute - + + + as an example of at work.

@@ -705,7 +708,7 @@ Linear Transformation Defined on a Basis -

Suppose $B=\set{\vectorlist{u}{n}}$ is a basis for the vector space $U$ and $\vectorlist{v}{n}$ is a list of vectors from the vector space $V$ (which are not necessarily distinct). Then there is a unique linear transformation, $\ltdefn{T}{U}{V}$, such that $\lt{T}{\vect{u}_i}=\vect{v}_i$, $1\leq i\leq n$.

+

Suppose $U$ is a vector space with basis $B=\set{\vectorlist{u}{n}}$ and the vector space $V$ contains the vectors $\vectorlist{v}{n}$ (which may not be distinct). Then there is a unique linear transformation, $\ltdefn{T}{U}{V}$, such that $\lt{T}{\vect{u}_i}=\vect{v}_i$, $1\leq i\leq n$.

@@ -729,35 +732,29 @@ \lt{T}{ a_1\vect{u}_1+ -a_2\vect{u}_2+ \cdots+ a_n\vect{u}_n+ b_1\vect{u}_1+ -b_2\vect{u}_2+ \cdots+ b_n\vect{u}_n }\\ \lt{T}{ \left(a_1+b_1\right)\vect{u}_1+ -\left(a_2+b_2\right)\vect{u}_2+ \cdots+ \left(a_n+b_n\right)\vect{u}_n } \text{}\\ \left(a_1+b_1\right)\vect{v}_1+ -\left(a_2+b_2\right)\vect{v}_2+ \cdots+ \left(a_n+b_n\right)\vect{v}_n a_1\vect{v}_1+ -a_2\vect{v}_2+ \cdots+ a_n\vect{v}_n+ b_1\vect{v}_1+ -b_2\vect{v}_2+ \cdots+ b_n\vect{v}_n \text{}\\ @@ -1277,16 +1274,11 @@

Then by , we have - -\lt{(T+S)}{\colvector{x_1\\x_2}} -= -\lt{T}{\colvector{x_1\\x_2}}+\lt{S}{\colvector{x_1\\x_2}} -= -\colvector{x_1+2x_2\\3x_1-4x_2\\5x_1+2x_2}+ -\colvector{4x_1-x_2\\x_1+3x_2\\-7x_1+5x_2} -= -\colvector{5x_1+x_2\\4x_1-x_2\\-2x_1+7x_2} - + + + +=\colvector{5x_1+x_2\\4x_1-x_2\\-2x_1+7x_2} + and by we know $T+S$ is also a linear transformation from $\complex{2}$ to $\complex{3}$.

@@ -1344,15 +1336,12 @@

For the sake of an example, choose $\alpha=2$, so by , we have - -\lt{\alpha T}{\colvector{x_1\\x_2\\x_3\\x_4}} -= -2\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4}} -= -2\colvector{x_1+2x_2-x_3+2x_4\\x_1+5x_2-3x_3+x_4\\-2x_1+3x_2-4x_3+2x_4} -= -\colvector{2x_1+4x_2-2x_3+4x_4\\2x_1+10x_2-6x_3+2x_4\\-4x_1+6x_2-8x_3+4x_4} - + + + + + + and by we know $2T$ is also a linear transformation from $\complex{4}$ to $\complex{3}$.

@@ -1428,8 +1417,8 @@

Then by -\lt{S}{\lt{T}{\colvector{x_1\\x_2}}}\\ - +\lt{S}{\lt{T}{\colvector{x_1\\x_2}}} + 2(x_1+2x_2)-(3x_1-4x_2)+(5x_1+2x_2)-(6x_1-3x_2)\\ 5(x_1+2x_2)-3(3x_1-4x_2)+8(5x_1+2x_2)-2(6x_1-3x_2)\\ @@ -1610,7 +1599,7 @@ -Find the matrix representation of $\ltdefn{T}{\complex{3}}{\complex{4}}$ given by +Find the matrix representation of $\ltdefn{T}{\complex{3}}{\complex{4}}$, $\lt{T}{\colvector{x\\y\\z}} = \colvector{3x + 2y + z\\ x + y + z \\ x - 3y \\2x + 3y + z }$. diff --git a/src/section-SLT.xml b/src/section-SLT.xml --- a/src/section-SLT.xml +++ b/src/section-SLT.xml @@ -539,7 +539,7 @@ \rng{T}=\spn{}

-

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 +

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 \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} @@ -705,9 +705,13 @@ Spanning Set for Range of a Linear Transformation -

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $S=\set{\vectorlist{u}{t}}$ spans $U$. Then +

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and -R=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}} + + +spans $U$. Then + + spans $\rng{T}$.

@@ -818,7 +822,15 @@ Surjective Linear Transformations and Bases -

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $B=\set{\vectorlist{u}{m}}$ is a basis of $U$. Then $T$ is surjective if and only if $C=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}$ is a spanning set for $V$.

+

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and + + + +is a basis of $U$. Then $T$ is surjective if and only if + + + +is a spanning set for $V$.