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\begin{document}

\noindent \hfill
\fbox{\fbox{\parbox{7.25in}{\scshape{Math 5230 \hfill Example \hfill Coordinate Matrices}}} \hfill }

\vspace{0.3in}

\noindent Let's explore coordinate representations of matrices and changes of basis.

\vspace{0.1in}

Let $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined by 
$T(x,y) = (3x+2y,4y)$. We will consider the relationships between matrix representations
of $T$ in coordinate systems associated with the following bases:
\begin{itemize}
   \item $\alpha = \{ (1,0), (0,1) \}$ (The standard basis)
   \item $\beta = \{ (1,1), (-2,3) \}$
   \item $\beta' = \{ (1,-1), (0,1) \}$
\end{itemize}

\noindent
First, let's compute $[T]_\alpha$ (the standard matrix of $T$), $[T]_\beta$, 
and $[T]_{\beta'}$.

\begin{itemize}
   \item $T((1,0)) = (3,0) = 3(1,0) + 0(0,1)$ so that $\displaystyle{[T((1,0))]_\alpha = \begin{bmatrix} 3 \\ 0 \end{bmatrix}}$ and
   
         $T((0,1)) = (2,4) = 2(1,0) + 4(0,1)$ so that $\displaystyle{[T((0,1))]_\alpha = \begin{bmatrix} 2 \\ 4 \end{bmatrix}}$. Therefore, 
\[ [T]_\alpha = \left[ \begin{array}{cc} [T((1,0))]_\alpha & [T((0,1))]_\alpha \end{array} \right] = \begin{bmatrix} 3 & 2 \\ 0 & 4 \end{bmatrix} = A \] 

   \item $\displaystyle{T((1,1)) = (5,4) = \frac{23}{5}(1,1) - \frac{1}{5}(-2,3)}$ so that
         $\displaystyle{[T((1,1))]_\beta = \begin{bmatrix} 23/5 \\ -1/5 \end{bmatrix}}$ and 
         
         $\displaystyle{T((-2,3)) = (0,12) = \frac{24}{5}(1,1) + \frac{12}{5}(-2,3)}$ so that
         $\displaystyle{[T((-2,3))]_\beta = \begin{bmatrix} 24/5 \\ 12/5 \end{bmatrix}}$. Therefore,
\[ [T]_\beta = \left[ \begin{array}{cc} [T((1,1))]_\beta & [T((-2,3))]_\beta \end{array} \right] = \frac{1}{5} \begin{bmatrix} 23 & 24 \\ -1 & 12 \end{bmatrix} = B \] 

   \item $T((1,-1)) = (1,-4) = 1(1,-1)-3(0,1)$ so that 
         $\displaystyle{[T((1,-1))]_{\beta'} = \begin{bmatrix} 1 \\ -3 \end{bmatrix}}$ and \\
         $T((0,1)) = (2,4) = 2(1,-1)+6(0,1)$ so that 
         $\displaystyle{[T((1,-1))]_{\beta'} = \begin{bmatrix} 2 \\ 6 \end{bmatrix}}$. Therefore,
\[ [T]_{\beta'} = \left[ \begin{array}{cc} [T((1,-1))]_{\beta'} & [T((0,1))]_{\beta'} \end{array} \right] = \begin{bmatrix} 1 & 2 \\ -3 & 6 \end{bmatrix} = C \] 
\end{itemize}

\noindent
Next, let's compute change of basis matrices. To convert from $\beta$-coordinates 
to $\alpha$-coordinates and then from $\beta'$-coordinates to $\alpha$-coordinates.
\begin{itemize}
   \item $(1,1) = 1(1,0) + 1(0,1)$ so that 
         $\displaystyle{[(1,1)]_\alpha = \begin{bmatrix} 1 \\ 1 \end{bmatrix}}$ and \\
         $(-2,3) = -2(1,0) + 3(0,1)$ so that 
         $\displaystyle{[(-2,3)]_\alpha = \begin{bmatrix} -2 \\ 3 \end{bmatrix}}$. Therefore, we can change from $\beta$ to $\alpha$ coordinates with the following matrix:
\[ [I]_{\beta}^{\alpha} = \begin{bmatrix} 1 & -2 \\ 1 & 3 \end{bmatrix} = P \]

   \item $(1,-1) = 1(1,0) -1(0,1)$ so that
         $\displaystyle{[(1,-1)]_\alpha = \begin{bmatrix} 1 \\ -1 \end{bmatrix}}$ and \\
         $(0,1) =  0(1,0) + 1(0,1)$ so that 
         $\displaystyle{[(0,1)]_\alpha = \begin{bmatrix} 0 \\ 1 \end{bmatrix}}$. Therefore, we can change from $\beta'$ to $\alpha$ coordinates with the following matrix:
\[ [I]_{\beta'}^{\alpha} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} = Q \]

   \item To change from $\beta$ to $\beta'$ coordinates we can follow: 
         $\beta \rightarrow \alpha \rightarrow \beta'$ thus
\[ [I]_\beta^{\beta'} = [I]_\alpha^{\beta'} [I]_\beta^{\alpha} = 
   Q^{-1}P = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \]
         (Or directly, $(1,1) = 1(1,-1)+2(0,1)$ and $(-2,3) = -2(1,-1)+1(0,1)$.)
\end{itemize}

Notice the following:
\[ [T]_\beta = [I]_\alpha^\beta [T]_\alpha [I]_\beta^\alpha = P^{-1}AP = B \]
\[ [T]_{\beta'} = [I]_\alpha^{\beta'} [T]_\alpha [I]_{\beta'}^\alpha = Q^{-1}AQ = C \]
\[ [T]_\beta^{\beta'} = [I]_\beta^{\beta'} [T]_\beta = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \frac{1}{5} \begin{bmatrix} 23 & 24 \\ -1 & 12 \end{bmatrix} = 
\begin{bmatrix} 5 & 0 \\ 9 & 12 \end{bmatrix} \]
or 
\[ [T]_\beta^{\beta'} = [I]_\alpha^{\beta'} [T]_\alpha [I]_\beta^\alpha = Q^{-1}AP =
\begin{bmatrix} 5 & 0 \\ 9 & 12 \end{bmatrix} \]

Finally, let ${\bf v} = (-1,3)$ then 
$\displaystyle{[{\bf v}]_\alpha = \begin{bmatrix} -1 \\ 3 \end{bmatrix}}$
so that $\displaystyle{[{\bf v}]_{\beta'} = [I]_\alpha^{\beta'}[{\bf v}]_\alpha =
Q^{-1}\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}}$.

\vspace{0.05in}

Also, $\displaystyle{[{\bf v}]_{\beta} = [I]_\alpha^{\beta}[{\bf v}]_\alpha =
P^{-1}\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix}}$.
\[ [T({\bf v})]_{\beta'} = [T]_\beta^{\beta'} [v]_\beta = 
\begin{bmatrix} 5 & 0 \\ 9 & 12 \end{bmatrix} \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix} = 
\begin{bmatrix} 3 \\ 15 \end{bmatrix} \]
On the other hand, $T({\bf v}) = T((-1,3)) = (3,12)$ and $\displaystyle{[T({\bf v})]_{\beta'} = [I]_\alpha^{\beta'} \begin{bmatrix} 3 \\ 12 \end{bmatrix} = 
Q^{-1} \begin{bmatrix} 3 \\ 12 \end{bmatrix} = \begin{bmatrix} 3 \\ 15 \end{bmatrix}}$.
\end{document}
