Matrix representation problem Consider the next linear operator:
$$T:M_{2\times2}(\mathbb{R})\longrightarrow M_{2\times2}(\mathbb{R})$$
such that $T\left(\begin{array}{cc}a&b\\c&d\end{array}\right)=\left(\begin{array}{cc}0&a+b\\0&0\end{array}\right)$. I need find the $Im(T)$ and the $Ker(T)$ but I don't how is the matrix representation of these linear operator, I've already calculate the images of the canonical vector basis:
$$T\left(\begin{array}{cc}1&0\\0&0\end{array}\right)=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$
$$T\left(\begin{array}{cc}0&1\\0&0\end{array}\right)=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$$
$$T\left(\begin{array}{cc}0&0\\1&0\end{array}\right)=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)$$
$$T\left(\begin{array}{cc}0&0\\0&1\end{array}\right)=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)$$
should I use a block matrix for the representation of $T$? I need help please.
An example of my work, if use the netx representation (I don't know if is it right):
$$[T]_{\beta}=\left(\begin{array}{cc|cc}0&1&0&0\\0&0&0&0\\\hline0&1&0&0\\ 0&0&0&0\end{array}\right)$$
making elemental operations by rows I have:
$$[T]_{\beta}=\left(\begin{array}{cc|cc}0&1&0&0\\0&0&0&0\\\hline0&0&0&0\\ 0&0&0&0\end{array}\right)$$
then we obtain $(a,b,c,d)=(a,0,c,d)$, in matrix form:
$$\left(\begin{array}{cc}a&0\\c&d\end{array}\right)=a\left(\begin{array}{cc}1&0\\0&0\end{array}\right)+c\left(\begin{array}{cc}0&0\\1&0\end{array}\right)+d\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$$
 A: The image of $T$ are all matrices of the form 
$$
\begin{bmatrix} 0&x\\ 0&0\end{bmatrix},\ \ \ x\in\mathbb R.
$$
The kernel will be those matrices mapped to zero, so we need $a+b=0$; that is, the kernel consists of those matrices of the form
$$
\begin{bmatrix} a&-a\\ c&d\end{bmatrix},\ \ \ \ a,c,d\in \mathbb R. 
$$
A: We note that the vector space $M_{2\times2}(\mathbb{R})$ has dimension $4$, since we have the following basis for the vector space:
$$ 
\mathbf{v}_1 = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} \qquad 
\mathbf{v}_2 = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} \qquad
\mathbf{v}_3 = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix} \qquad 
\mathbf{v}_4 = \begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} $$
We have that the transformation $T$ works as follows:
\begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix} &\xrightarrow[T]{} 
\begin{pmatrix} 0 & a+b \\ 0 & 0 \end{pmatrix} \\
T(a \mathbf{v}_1 + b\mathbf{v}_2 + c\mathbf{v}_3 + d\mathbf{v}_4) &= (a+b)\mathbf{v}_2 \\
A \begin{bmatrix} a \\ b \\ c \\ d\end{bmatrix} &= \begin{bmatrix}0 \\ a+b 
 \\ 0 \\ 0\end{bmatrix} \\[8pt]
A &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
\end{align*}
This is the matrix representation of the linear operator $T$.
The kernel of $T$ is the set of all elements that get mapped to the identity element $\mathbf{0} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.
\begin{align*}
\ker T &= \left\{ \begin{pmatrix} a & -a \\ c & d \end{pmatrix} \bigg\rvert \,\,\, a,c,d \in \mathbb{R} \right\} \\
&= \left\{A = [a_{ij}] \in M_{2\times 2}(\mathbb{R}) \mid a_{11} + a_{12} = 0 \right\}
\end{align*}
The image of $T$ is the set of all elements that can result from the transformation $T$, which is:
\begin{align*}
\text{im } T &= \left\{ \begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix} \bigg\rvert \,\,\, b \in \mathbb{R} \right\} \\
&= \left\{A = [a_{ij}] \in M_{2\times 2}(\mathbb{R}) \mid a_{11} = a_{21} = a_{22} = 0 \right\}
\end{align*}
Another way to find the kernel and image of $T$ is to find the nullspace (kernel) and column space (image) of $A$.
