Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given 
$$ A = \begin{pmatrix} 
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t X^t
\end{matrix}$$
Let $B$ be the standard basis for $\mathbb{R}^{2 \times 2}$ :
$$B =\left\{ \begin{pmatrix} 
1 & 0\\
0 & 0 
\end{pmatrix}, \begin{pmatrix} 
0 & 1\\
0 & 0\\ 
\end{pmatrix},\begin{pmatrix} 
0 & 0\\
1 & 0\\ 
\end{pmatrix},\begin{pmatrix} 
0 & 0\\
0 & 1\\ 
\end{pmatrix} \right\}$$
Calculate $\textsf{M}_B(\varphi)$ we come to $$\textsf{M}_B(\varphi) = \begin{pmatrix} 
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
We calculate a basis for the kernel like this:
If 
$$X:= \begin{pmatrix} 
a & b\\
c & d\\ 
\end{pmatrix}$$
then $$\varphi(X) = \begin{pmatrix} 
a-b & -a+b\\
c-d & -c+d\\ 
\end{pmatrix}+\begin{pmatrix} 
a-b & c-d\\
-a+b & -c+d\\ 
\end{pmatrix} = \begin{pmatrix} 
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\ 
\end{pmatrix}$$
Now we have to look, for what values $$\begin{pmatrix} 
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\ 
\end{pmatrix}$$
by definition, the kernel of a linear transformation is $\varphi(X) = 0$, therefore our basis for $\ker(\varphi)$ should be 
$$\left\{ \begin{pmatrix} 1 & 1 \\ 0 & 0  \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1  \end{pmatrix} \right\}$$
Now here comes the part where I'm confused. How do I calculate a basis for $\operatorname{im}(\varphi)$ ?
$\textsf{M}_B(\varphi)$ is the transformation matrix. I've read that you'd just transpose the matrix $\textsf{M}_B(\varphi)$ and row reduce to calculate a basis. I just don't get it.
The solution according to the solution it should be the basis $$ \left\{\begin{pmatrix}
0 & 1 \\
1 & -2  \end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & -1  \end{pmatrix} \right\}$$
Also little side questions. I do know about 
$$\dim(A) = \dim(\operatorname{im} A) + \dim(\ker A)$$
but how exactly do you know the dimension of the Kernel/Image?

I think if you take your linear map $\varphi$ and write it as a matrix in terms of the basis you provided, then it becomes
$$\begin{pmatrix}2 & -2&0 &0\\ -1 &1&1&-1\\-1&1&1&-1\\ 0&0&-2&2\end{pmatrix}$$
The column space of this linear map is spanned by two vectors, namely $(2,-1,-1,0)$ and $(0,1,1,-2)$. Rewriting these as matrices will give you your answer.
This also gives you an easier method for finding the basis of the kernel, by row reducing this matrix and studying the resulting linear equations.
