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?

Observe that $(XA+A^TX^T)$ is a symmetric matrix. So the image will be a subset of the set of symmetric matrices. But the question about a basis for the image set can be resolved as follows:
$$\varphi(X)=\begin{bmatrix} 
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\ 
\end{bmatrix}=(a-b)\begin{bmatrix}2&-1\\-1&0\end{bmatrix}+(c-d)\begin{bmatrix}0&1\\1&-2\end{bmatrix}.$$
This means that anything that belongs to the image is in the span of the two matrices on the RHS. Observe that both the matrices on the right side are LI. Hence they form a basis for the image set. Furthermore the dimension of the image set is $2$ (because we have $2$ vectors in the basis).
NOTE:
The answer given in your textbook is not different from what I gave above because $$\begin{bmatrix}2&-1\\-1&0\end{bmatrix}+\begin{bmatrix}0&1\\1&-2\end{bmatrix}=2\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$ 
