<s>[INST] How to find basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T$? 
Find the basis of $\ker T$ and $\mathrm{Im} T$ for the linear map $T:M^{\mathbb R}_{2 \times 2} \to M^{\mathbb R}_{2 \times 2}$ defined as $T(A)=A-A^t$ for all $A \in M^{\mathbb R}_{2 \times 2}$.

Let $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$. Then:
$$
T(A)=\begin{pmatrix} a&b\\c&d \end{pmatrix}-\begin{pmatrix} a&c\\b&d \end{pmatrix}=\begin{pmatrix} 0&b-c\\c-b&0 \end{pmatrix}\stackrel{R_2 \gets-1\cdot R_2}{=}\begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix}
$$
In order to find $\ker T$ we'll evalute $T(A)=0$:
$$
\begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix}=\begin{pmatrix} 0&0\\0&0 \end{pmatrix}\Rightarrow b=c \Rightarrow \ker T=span \Biggl\{ \begin{pmatrix} 1&0\\0&0 \end{pmatrix}, \begin{pmatrix} 0&1\\1&0 \end{pmatrix},\begin{pmatrix} 0&0\\0&1 \end{pmatrix} \Biggr\}
$$
Thus $\dim(\ker T)=3$.

Regarding $\mathrm{Im} T$:
$$
ImT=\Biggl\{ \begin{pmatrix} 0&b-c\\b-c&0 \end{pmatrix} \text{such that}\quad b,c, \in \mathbb R \Biggr\}=span\Biggl\{ \begin{pmatrix} 0&1\\1&0 \end{pmatrix} \Biggr\}
$$
Thus $\dim(\mathrm{Im} T)=1$.
Are my calculations correct?
 [/INST] You can use the fact that every square matrix is, in a unique way, the sum of a symmetric and an antisymmetric matrices.
Any symmetric matrix belongs to the kernel of $T$, whereas the antisymmetric matrix
$$
A=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$
does not belong to $\ker T$. Since the space of symmetric matrices has dimension $3$, with basis
$$
\left\{
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\right\}
$$
this is also a basis for $\ker T$.
A basis of the image consists of
$$
T(A)=\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}
$$
</s>