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\noindent \hfill
\fbox{\fbox{\parbox{7.25in}{\scshape{Math 5230 \hfill Linear Transformations \hfill Composition, Kernels, and Ranges}}} \hfill }

\vspace{0.15in}

\noindent Consider the following linear transformations:

\vspace{-0.2in}

\[ S:P_2 \rightarrow \mathbb{R}^{2 \times 2} \qquad \mathrm{defined\;by} \qquad 
   S(ct^2+bt+a) = \begin{bmatrix} a+b & b \\ c & a+c \end{bmatrix} \]

\vspace{-0.1in}

\[ T:\mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}^2 \qquad \mathrm{defined\;by} \qquad
   T\left(\begin{bmatrix} x & y \\ u & v \end{bmatrix} \right) = (x-y,u-v) \]

\noindent
Notice that if we compose these maps we get $T \comp S:P_2 \rightarrow \mathbb{R}^2$
where

\vspace{-0.2in}

$$
 (T \comp S)(ct^2+bt+a)  = T(S(ct^2+bt+a)) = T\left(\begin{bmatrix} a+b & b \\ c & a+c \end{bmatrix}\right)
                         = (a+b-b,c-(a+c)) = (a,-a) 
$$

\vspace{-0.05in}

\noindent Consider the standard bases: \qquad $\beta = \{1,t,t^2\}$ for $P_2$, \qquad  $\delta = \{ e_1 = (1,0), e_2 = (0,1) \}$ for $\mathbb{R}^2$, \quad and

\vspace{0.1in}

\noindent
   $\displaystyle{\gamma = \left\{ 
          E_{11} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, 
          E_{12} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, 
          E_{21} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, 
          E_{22} = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right\}}$ for $\mathbb{R}^{2 \times 2}$.

\vspace{0.1in}  

\noindent Let's find coordinate matrices for $S$, $T$, and $T \circ S$.

\vspace{-0.3in}

\begin{itemize}
   \item $\displaystyle{ S(1) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 
    1E_{11}+0E_{12}+0E_{21}+1E_{22} \qquad \Longrightarrow \quad [S(1)]_\gamma = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}}$

\vspace{-0.2in}

    $\displaystyle{ \qquad S(t) = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} = 
    1E_{11}+1E_{12}+0E_{21}+0E_{22} \qquad \Longrightarrow \qquad [S(1)]_\gamma = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}}$

\vspace{-0.2in}

    $\displaystyle{ S(t^2) = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} = 
    0E_{11}+0E_{12}+1E_{21}+1E_{22} \qquad \Longrightarrow \quad [S(1)]_\gamma = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}}$

\vspace{-0.25in}

    $\displaystyle{\qquad \qquad \mathrm{Therefore,} \qquad [S]_\beta^\gamma = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix} }$

\vspace{-0.1in}
  
   \item $\displaystyle{ T(E_{11}) = T\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right) = (1,0) = 1e_1+0e_2
          \qquad \Longrightarrow \qquad [T(E_{11})]_\delta = \begin{bmatrix} 1 \\ 0 \end{bmatrix}}$
 
         $\displaystyle{ T(E_{12}) = T\left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right) = (-1,0) = -1e_1+0e_2
          \qquad \Longrightarrow \qquad [T(E_{12})]_\delta = \begin{bmatrix} -1 \\ 0 \end{bmatrix}}$

         $\displaystyle{ T(E_{21}) = T\left(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\right) = (0,1) = 0e_1+1e_2
          \qquad \Longrightarrow \qquad [T(E_{21})]_\delta = \begin{bmatrix} 0 \\ 1 \end{bmatrix}}$

         $\displaystyle{ T(E_{22}) = T\left(\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right) = (0,-1) = 0e_1-1e_2
          \qquad \Longrightarrow \qquad [T(E_{22})]_\delta = \begin{bmatrix} 0 \\ -1 \end{bmatrix}}$   

   $\displaystyle{\qquad\qquad \mathrm{Therefore,} \qquad
    [T]_\gamma^\delta = \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix}}$

   \item To find a coordinate matrix for $T \comp S$ we could do a direct computation like before...or we can use our work from the last two bullets:

\vspace{-0.35in}

\[ [T \comp S]_\beta^\delta = [T]_\gamma^\delta [S]_\beta^\gamma = 
   \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 \end{bmatrix} 
   \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{bmatrix} =
   \begin{bmatrix} 1 & 0 & 0 \\ -1 & 0 & 0 \end{bmatrix} \]
         [Or by direct computation:
         $(T \comp S)(1) = (1,-1) = 1e_1-1e_2$ and
         $(T \comp S)(t^2) = (T \comp S)(t) = (0,0) = 0e_1+0e_2$
         which gives us the same matrix (of course).] 
\end{itemize}

\vspace{-0.1in}

\noindent Let's find a basis for the Kernel and Range of $S$, $T$, and $T \circ S$.
          We know that if $X$ is a linear transformation with corresponding matrix $Y$ then
          $\mathrm{N}(Y)$ is a coordinate representation of $\mathrm{Ker}(X)$ and $\mathrm{Col}(Y)$ is a
          coordinate representation of $\mathrm{Range}(X)$.
\begin{itemize}
   \item For $S$ we have... 

\vspace{-0.65in}

         \[ [S]_\beta^\gamma = 
         \begin{bmatrix} 1 & 1 & 0 \\ 
                         0 & 1 & 0 \\ 
                         0 & 0 & 1 \\ 
                         1 & 0 & 1 \end{bmatrix} \sim
         \begin{bmatrix} 1 & 0 & 0 \\ 
                         0 & 1 & 0 \\ 
                         0 & 0 & 1 \\ 
                         0 & 0 & 0 \end{bmatrix} \]

\vspace{-0.25in}

        Thus $\displaystyle{ \mathrm{N}([S]_\beta^\gamma) = 
        \left\{ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \right\}}$ and so $\mathrm{Ker}(S) = \{ 0 \}$ which means $S$ is 1-1 and $\mathrm{nullity}(S)=0$. 

\vspace{-0.15in}

        Next, we see that every column of the coordinate matrix is a pivot column so
        that $\displaystyle{ \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix},
              \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} \right\}}$
        is a basis for $\mathrm{Col}([S]_\beta^\gamma)$. These coordinate vectors correspond to 

\vspace{-0.1in}

the following set (which is a basis for $\mathrm{Range}(S)$): $\displaystyle{ \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},
              \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \right\}}$
        which is a basis for $\mathrm{Range}(S)$. Thus $\mathrm{rank}(S)=3$ (obviously $S$ is not onto). 

    \item For $T$ we have...

\vspace{-0.5in}

          \[ [T]_\gamma^\delta = 
          \begin{bmatrix} 1 & -1 & 0 &  0 \\ 
                          0 &  0 & 1 & -1 \end{bmatrix} \]
          Labeling variables $x_1$, $x_2$, $x_3$, and $x_4$, we have the equations: $x_1-x_2=0$ and $x_3-x_4=0$.
          $x_2$ and $x_4$ are free, so let $x_2=s$ and $x_4=t$ we get:

\vspace{-0.25in}

          \[ \qquad \qquad \begin{array}{ccc} x_1 & = & s \\ x_2 & = & s \\ x_3 & = & t \\ x_4 & = & t \end{array} \qquad
             \mathrm{so\;that...} \qquad {\bf x} = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}s + 
                                               \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}t \]

\vspace{-0.3in}

          Thus, $\displaystyle{\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, 
                              \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} \right\}}$ is a basis 
          for $\mathrm{N}([T]_\gamma^\delta)$ which corresponds to:
          $\displaystyle{ \left\{ \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, 
                                  \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \right\} }$ 
          (a basis for $\mathrm{Ker}(T)$). Therefore, $T$ is not 1-1 and $\mathrm{nullity}(T)=2$.

\vspace{-0.1in}

          Next, the first and third columns of our coordinate matrix are pivot columns so that 
          $\displaystyle{\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right \}}$
          is a basis for $\mathrm{Col}([T]_\gamma^\delta)$. These coordinate vectors correspond to 
          $\{ (1,0), (0,1) \}$ (the standard basis for $\mathbb{R}^2$). Therefore,
          $\mathrm{Range}(T)=\mathbb{R}^2$ and $\mathrm{rank}(T)=2$.

   \item Finally, for $T \comp S$ we have...

\vspace{-0.5in}

         \[ \qquad \qquad [T \comp S]_\beta^\delta =
         \begin{bmatrix}  1 & 0 & 0 \\ 
                         -1 & 0 & 0 \end{bmatrix} \sim 
         \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
         If we label variables $x_1$, $x_2$, and $x_3$, we see that the matrix says: $x_1=0$. Thus 
         $x_2$ and $x_3$ are free, say $x_2=s$ and $x_3=t$ so we get: 

\vspace{-0.35in}

          \[ \qquad \qquad \qquad \begin{array}{ccc} x_1 & = & 0 \\ x_2 & = & s \\ x_3 & = & t \end{array} \qquad
             \mathrm{thus...} \qquad {\bf x} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}s + 
                                               \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}t \]

\vspace{-0.3in}

          Therefore, $\displaystyle{\left\{ \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, 
                              \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}}$ is a basis 
          for $\mathrm{N}([T \comp S]_\beta^\delta)$. These coordinate vectors correspond to:
          $\{ t, t^2 \}$ which is a basis for $\mathrm{Ker}(T \comp S)$. From this we see that $T \comp S$ is not 
          1-1 and $\mathrm{nullity}(T \comp S)=2$.

          Next, the first column of our coordinate matrix is the only pivot column so that 
          $\displaystyle{\left\{ \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\}}$
          is a basis for $\mathrm{Col}([T \comp S]_\beta^\delta)$. This coordinate vector corresponds to 
          $\{ (1,-1) \}$ which is a basis for $\mathrm{Range}(T \comp S)$. We see from this that $T \comp S$ is
          not onto since $\mathrm{rank}(T \comp S) = 1 \;(<2)$.
\end{itemize}

\end{document}
