<s>[INST] Finding $Ker(T_a)$ and $Im(T_a)$ there will be a matrix $A \in R^{2 x 2}$, $T_A:R^{2 x 2} \rightarrow R^{2 x 2}$
Define by $T_A(X)=$$
        \begin{pmatrix}
        1 & 2\\
        0 & 3 \\
        \end{pmatrix}
$$X-X$$
        \begin{pmatrix}
        1 & 2\\
        0 & 3 \\
        \end{pmatrix}
$$\forall X \in R^{2 x 2} $
Find $Ker(T_A)$ and it basis and $Im(T_A)$ and it basis.
As for $Ker(T_A)$ we need to find the matrix X so that $$
        \begin{pmatrix}
        1 & 2\\
        0 & 3 \\
        \end{pmatrix}X-X    \begin{pmatrix}
        1 & 2\\
        0 & 3 \\
        \end{pmatrix}=\begin{pmatrix}
        0 & 0\\
        0 & 0 \\
        \end{pmatrix}
$$
So I got the following equations $$\begin{cases} a_1+2a_3-a_1=0\\
a_2+2a_4-2a_1-3a_2=0\\
3a_3-a_3=0\\
3a_4-2a_3+3a_4=0
\end{cases}
$$
There is the trival answer that is 0 for all a, but the answer is not \begin{pmatrix}
        0 & 0\\
        0 & 0 \\
        \end{pmatrix}
 [/INST] If we denote 
$$X=\begin{pmatrix}
        a & b\\
        c & d \\
        \end{pmatrix}$$
then 
$$AX-XA=\begin{pmatrix}
        2c & (2d-2b-2a)\\
        2c & -2c \\
        \end{pmatrix}=\begin{pmatrix}
        0 & 0\\
        0 & 0 \\
        \end{pmatrix}\iff (c=0)\land (d=b+a)$$
hence
$$\operatorname{ker}(T_A)=\operatorname{span}\Bigg(I_2,\begin{pmatrix}
        0 & 1\\
        0 & 1 \\
        \end{pmatrix}\Bigg)$$
and clearly we have
$$\operatorname{Im}(T_A)=\operatorname{span}\Bigg(\begin{pmatrix}
        1 & 0\\
        1 & -1 \\
        \end{pmatrix},\begin{pmatrix}
        0 & 1\\
        0 & 0 \\
        \end{pmatrix}\Bigg)$$
</s>