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?
 A: I think an answer will help clarify things. You computed that $$T(A) = \begin{pmatrix} 0 & b-c \\ c-b & 0 \end{pmatrix} = \begin{pmatrix} 0 & b-c \\ -(b-c) & 0 \end{pmatrix} = (b-c) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_.$$
From this we see that $\textrm{Im}(T) = \textrm{span}\left\{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\right\}_.$
We also see that $T(A) = 0$ if and only if $b-c = 0$. In other words when $b=c$. From there you see that $$\ker(T) = \textrm{span}\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\}_.$$
Note you shouldn't actually be row reducing $T(A)$, you want that matrix to be zero. You could find the matrix of $T$ with respect to some basis, and row reduce that to find the kernel of $T$.
The reason that row reduction can be used to find the kernel of a matrix is because row reduction corresponds to multiplying on the left by an invertible matrix. This doesn't change the kernel.

An alternative approach to the problem. Let us compute the matrix of $T$, call it $B$, with respect to the standard basis of $M_{2\times 2}(\mathbb{R})$, $\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\}_.$
$$T\left(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}_. $$
$$T\left(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \right) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}_. $$
$$T\left(\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}_. $$
$$T\left(\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}_. $$
Using this, we calculate that $$ B = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & -1 & 0 \\ 0 & -1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}_.$$
You could row reduce this or just see by inspection that $A \in \ker (T)$ precisely when $b = c$.
A: 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
$$
T(A)=\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}
$$
