\documentclass[11pt]{article}

\usepackage{enumerate}
\usepackage{ifthen}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{color}

%\pagestyle{empty}
\topmargin -0.75in
\oddsidemargin -0.25in
\evensidemargin -0.25in
\textheight 9.25in
\textwidth 7.0in 

\definecolor{ignoreGray}{rgb}{0.75,0.75,0.75}
\newcommand{\ig}{\color{ignoreGray}}

\newcommand{\myspace}{\mbox{\parbox[c][0.35in]{0.35in}{\hfill}}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}

\newcommand{\B}[1]{{\bf #1}}

\newcommand{\E}[2]{\B e_{#1}\B e_{#2}^T}

\newcommand{\force}{{$\mbox{\;}$}}

\begin{document}

\noindent \hfill
\fbox{\fbox{\parbox{6.9in}{{\scshape{Linear Algebra \hfill Coordinates vs. Coordinate Matrix}} }}} \hfill 

\vspace{0.15in}

\noindent Let's explore the following linear transformation:

\vspace{-0.15in}

$$T:\mathbb{R}^{2 \times 2} \to P_1 = \{ax+b \;|\; a,b \in \mathbb{R}\} \qquad \mbox{defined by} \qquad T\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = (a+2b-c)x+(a+d)$$

\noindent {\bf\Large $T$ is linear:} Let's prove $T$ preserves addition and scalar multiplication. 

$$T\left(\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}\right) = T\left(\begin{bmatrix} a_1+a_2 & b_1+b_2 \\ c_1+c_2 & d_1+d_2 \end{bmatrix}\right) = ((a_1+a_2)+2(b_1+b_2)-(c_1+c_2))x+((a_1+a_2)+(d_1+d_2))$$
$$=\left((a_1+2b_1-c_1)x+(a_1+d_1)\right)+\left((a_2+2b_2-c_2)x+(a_2+d_2)\right) = T\left(\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}\right)  + T\left(\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}\right)$$

$$T\left(s\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = T\left(\begin{bmatrix} sa & sb \\ sc & sd \end{bmatrix}\right) = (sa+2sb-sc)x+(sa+sd) = s((a+2b-c)x+(a+d)) =  s\;T\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) $$

\noindent {\bf\Large Standard coordinate matrix:} Let $\alpha = \{E_{11},E_{12},E_{21},E_{22}\}$ and $\beta = \{1,x\}$. 

These are the standard bases for $\mathbb{R}^{2 \times 2}$ and $P_1$. 

To find the coordinate matrix we plug each $\alpha$ (input basis) vector into our map:
$T(E_{11}) = (1+2(0)-0)x+(1+0)=x+1$,  $T(E_{12}) = (0+2(1)-0)x+(0+0)=2x$, $T(E_{21}) = (0+2(0)-1)x+(0+0)=-x$, $T(E_{22}) = (0+2(0)-0)x+(0+1)=1$. Then we write these in terms of $\beta$ (output basis) coordinates (note the order of $\beta$ -- constant term then coefficient of $x$):
$$[T]_\alpha^\beta = \left[ \; [T(E_{11})]_\beta \  [T(E_{12})]_\beta \  [T(E_{21})]_\beta \  [T(E_{22})]_\beta \; \LARGE\strut \right]  = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 1 & 2 & -1 & 0 \end{bmatrix}$$

\noindent {\bf\Large Standard basis of linear transformations:} Keep the same bases $\alpha$ and $\beta$. The map that sends the $j$-th $\alpha$ (input basis) vector to the $i$-th $\beta$ (output basis) vector  and all other input vectors to $\B 0$ is called $T_{ij}$. In particular,

$$T_{11}(E_{11})=1, \quad T_{11}(E_{12})=0, \quad T_{11}(E_{21})=0, \quad \mbox{and} \quad T_{11}(E_{22})=0$$
$$T_{12}(E_{11})=0, \quad T_{12}(E_{12})=1, \quad T_{12}(E_{21})=0, \quad \mbox{and} \quad T_{12}(E_{22})=0$$
$$T_{13}(E_{11})=0, \quad T_{13}(E_{12})=0, \quad T_{13}(E_{21})=1, \quad \mbox{and} \quad T_{13}(E_{22})=0$$
$$T_{14}(E_{11})=0, \quad T_{14}(E_{12})=0, \quad T_{14}(E_{21})=0, \quad \mbox{and} \quad T_{14}(E_{22})=1$$
$$T_{21}(E_{11})=x, \quad T_{21}(E_{12})=0, \quad T_{21}(E_{21})=0, \quad \mbox{and} \quad T_{21}(E_{22})=0$$
$$T_{22}(E_{11})=0, \quad T_{22}(E_{12})=x, \quad T_{22}(E_{21})=0, \quad \mbox{and} \quad T_{22}(E_{22})=0$$
$$T_{23}(E_{11})=0, \quad T_{23}(E_{12})=0, \quad T_{23}(E_{21})=x, \quad \mbox{and} \quad T_{23}(E_{22})=0$$
$$T_{24}(E_{11})=0, \quad T_{24}(E_{12})=0, \quad T_{24}(E_{21})=0, \quad \mbox{and} \quad T_{24}(E_{22})=x$$

These maps are rigged up so that $[T_{ij}]_\alpha^\beta = E_{ij}$. So just as $\{E_{11},E_{12},E_{13},E_{14},E_{21},E_{22},E_{23},E_{24}\}$ is a basis for $\mathbb{R}^{2 \times 4}$, we have that $\gamma = \{T_{11},T_{12},T_{13},T_{14},T_{21},T_{22},T_{23},T_{24}\}$ is a basis for the space of linear transformations from $\mathbb{R}^{2 \times 4}$ to $P_1$ (that is  $\mathcal{L}(\mathbb{R}^{2 \times 2}, P_1) = \mathrm{Hom}_\mathbb{R} (\mathbb{R}^{2 \times 2}, P_1)$).

For example, notice that $(T_{11}+T_{14}+T_{21}+2T_{22}-T_{23})(E_{11}) = T_{11}(E_{11})+T_{14}(E_{11})+T_{21}(E_{11})+2T_{22}(E_{11})-T_{23}(E_{11}) = 1+0+x+2(0)-0 = 1+x = T(E_{11})$. In fact, $T_{11}+T_{14}+T_{21}+2T_{22}-T_{23}$ and $T$ match on all the $\alpha$ basis vectors. Therefore, $T=T_{11}+T_{14}+T_{21}+2T_{22}-T_{23}$. 

Our coordinate isomorphisms are compatible with this as well. Notice that $[T]_\alpha^\beta = E_{11}+E_{14}+E_{21}+2E_{22}-E_{23}$ (changing the $T_{ij}$'s to $E_{ij}'s$). 

Therefore, in $\gamma$ coordinates we have\dots

$$T \mbox{ as a vector in }\mathcal{L}(\mathbb{R}^{2 \times 2}, P_1): \quad [T]_\gamma = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ 2 \\ -1 \\ 0 \end{bmatrix}$$ 

\dots compared with\dots

$$T \mbox{ as a coordinate matrix:} \qquad \qquad [T]_\alpha^\beta = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 1 & 2 & -1 & 0 \end{bmatrix}$$

\noindent {\bf\Large In general:} Let $V$ be an $n$-dimensional space (over $\mathbb{F}$) with basis $\alpha = \{ \B v_1,\dots,\B v_n \}$ and let $W$ be an $m$-dimensional space (over $\mathbb{F}$) with basis $\beta = \{\B w_1,\dots,\B w_m\}$. For each $1 \leq k \leq n$, define $$T_{ij}(\B v_k) = \delta_{jk}\B w_i = \left\{ \begin{array}{ccc} \B w_i & & j=k \\ \B 0 & & j \not= k \end{array} \right.$$ (this is the map that sends input basis vector $j$ to output basis vector $i$ and then kills the rest of the input vectors). 

We have that $\gamma = \{ T_{11},\dots,T_{1n},T_{21},\dots,T_{2n},\dots,T_{m1},\dots,T_{mn} \}$ is a basis for $ \mathcal{L}(V,W) = \mathrm{Hom}_\mathbb{F}(V,W)$ (the space of all linear maps from $V$ to $W$).

Let $T:V \to W$ be a linear map (i.e. $T \in \mathcal{L}(V,W) = \mathrm{Hom}_\mathbb{F}(V,W)$). Let $A=(a_{ij}) = [T]_\alpha^\beta$. This means that $T(\B v_k) = \sum\limits_{i=1}^m a_{ik}\B w_i$ (the $k$-th column of $A=[T]_\alpha^\beta$ is given by the coordinates of $T(\B v_k)$).

Consider $S = \sum\limits_{i=1}^m \sum\limits_{j=1}^n a_{ij}T_{ij}$. Then $S(\B v_k) = \sum\limits_{i=1}^m \sum\limits_{j=1}^n a_{ij}T_{ij}(\B v_k) =  \sum\limits_{i=1}^m \sum\limits_{j=1}^n a_{ij}\delta_{jk}\B w_i = \sum\limits_{i=1}^m a_{ik}\B w_i$ (only the $j=k$ terms survive). Therefore, $S(\B v_k)=T(\B v_k)$ for $1 \leq k \leq n$. Since $S$ and $T$ match on a basis, $S=T$. We have that $T = \sum\limits_{i=1}^m \sum\limits_{j=1}^n a_{ij}T_{ij}$. In other words, the $\gamma$-coordinates of $T$ are exactly the entries of $A = [T]_\alpha^\beta$ (its coordinate matrix).

$$ \mbox{If } \quad [T]_\alpha^\beta = A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \qquad \mbox{then} \qquad [T]_\gamma = \begin{bmatrix} a_{11} \\ \vdots \\ a_{1n} \\ a_{21} \\ \vdots \\ a_{2n} \\ \vdots \\ a_{m1} \\ \vdots \\ a_{mn} \end{bmatrix}.$$


\end{document}
