Basis for kernel and range for a linear transformation of polynomials mapped to the vector space of 2x2 Let $V$ be the vector space $P_2$(x) of polynomials in $x$ of degree 2 or
less, $W$ be the vector space $M_{2,2}$ of 2×2 real matrices and $T$ be the linear
transformation
$T: V → W:$ $a + bx + cx^2$ $\mapsto$ \begin{bmatrix} a-b & b-c \\ 0 & c-a \end{bmatrix}
Compute bases for the kernel and range of $T$ and for $V$ and hence verify the
general rank-nullity theorem for $T : V → W$.
I don't understand how to get the kernel and range when the linear transformation maps to a vector space of matrices and so cannot compute bases. Any help/solutions would be appreciated. 
 A: For the kernel of $T$ you want those polynomials in $V$ that map to the zero matrix in $W$. So
\begin{align*}
T(a+bx+cx^2) & = \begin{bmatrix}0&0\\0&0\end{bmatrix}\\
\begin{bmatrix} a-b & b-c \\ 0 & c-a \end{bmatrix} & = \begin{bmatrix}0&0\\0&0\end{bmatrix}.
\end{align*}
This gives $a=b=c$. So the polynomials which lie in the kernel are of the form $a(1+x+x^2)$, where $a \in \mathbb{R}$. So a basis for the kernel is
$$\mathcal{B}_{\text{ker}}=\{1+x+x^2\} \implies \dim(\text{Ker} T)=1$$ 
Likewise we can go for a basis for the range. First we can get the range: assume that $\begin{bmatrix}p&q\\r&s\end{bmatrix} \in \text{Range }(T)$, then there exists some polynomial $a+bx+cx^2 \in V$ such that 
\begin{align*}
T(a+bx+cx^2) & =\begin{bmatrix}p&q\\r&s\end{bmatrix}\\
\begin{bmatrix} a-b & b-c \\ 0 & c-a \end{bmatrix}& = \begin{bmatrix}p&q\\r&s\end{bmatrix}
\end{align*}
This gives the following system:
$$
\begin{align*}
a-b & = p\\
b-c & = q\\
0 & = r\\
c-a&=s
\end{align*}
 \Longrightarrow 
\begin{bmatrix}
1&-1&0&|&p\\
0&1&-1&|&q\\
0&0&0&|&r\\
-1&0&1&|&s
\end{bmatrix}
\Longrightarrow 
\begin{bmatrix}
1&-1&0&|&p\\
0&1&-1&|&q\\
0&0&0&|&r\\
0&0&0&|&p+q+s
\end{bmatrix}
$$
From this it follows that the range only consists of matrices of the form $\begin{bmatrix}p&q\\0&-p-q\end{bmatrix}$. Now we can go for a basis for the range of $T$ as follows:
$$\begin{bmatrix}p&q\\0&-p-q\end{bmatrix}=p\begin{bmatrix}1&0\\0&-1\end{bmatrix}+q\begin{bmatrix}0&1\\0&-1\end{bmatrix}$$
This shows that 
$$\mathcal{B}_{\text{range}}=\left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&-1\end{bmatrix}\right\} \implies \dim(\text{Range } T)=2$$ 
