Consider the vector space of $\mathbb{R}^{2\times 2}$ and the linear operator $L_{A,B}:$ $\mathbb{R}^{2\times 2} \to \mathbb{R}^{2\times 2}$ is given by $L_{A,B} = AX + XB$ where $A = \begin{bmatrix} 1&2\\ 3&4\\ \end{bmatrix} $ and $B = \begin{bmatrix} 8&7\\ 6&5\\ \end{bmatrix} $.
Find the matrix representation of $L_{a,b}$ with respect to the basis $\begin{bmatrix} 1&0\\ 0&0\\ \end{bmatrix} $ $\begin{bmatrix} 0&1\\ 0&0\\ \end{bmatrix} $ $\begin{bmatrix} 0&0\\ 1&0\\ \end{bmatrix} $ $\begin{bmatrix} 0&0\\ 0&1\\ \end{bmatrix} $ .
Normally if you want to map a vector into itself you place the answers next to each other I thought. But then you get a $4\times 4$ matrix.
Which would be
If you fill in all basis in as $X$ you get out $4$ matrices which you fill in as the point where
$$ \begin{bmatrix} 9&7&6&6\\ 3&0&0&3\\ 2&0&0&2\\ 12&7&6&9\\ \end{bmatrix} $$
And then if you would do Ax you wouldn't be able to perform the multiplication. I thought maybe you could add them but that seems weird to me because you normally don't do that.
Then you would get $\begin{bmatrix} 9&7\\ 3&0\\ \end{bmatrix} + \begin{bmatrix} 6&6\\ 0&3\\ \end{bmatrix} + \begin{bmatrix} 2&0\\ 12&7\\ \end{bmatrix} + \begin{bmatrix} 0&2\\ 6&9\\ \end{bmatrix} =\begin{bmatrix} 17&15\\ 21&19\\ \end{bmatrix} $
But it looks wrong because normally I don't have to add them but have to place them next to each other. But then we get the problem that we cant multiply it. Is there some Obvious other answer or am i just thinking wrong and it is just a $4\times 4$ matrix or adding them.
Your question is a little bit ambiguous but this is how I interpret it: You have a linear operator $L_{A,B}: \mathbb{R}^{2\times 2}\to \mathbb{R}^{2\times 2}$ given by $L_{A,B}(X) = AX + XB$ where $A$ and $B$ are the matrices in your question.
Note that your linear operator is from $\mathbb{R}^{2\times 2} \cong \mathbb{R}^4$ to $\mathbb{R}^{2\times 2} \cong \mathbb{R}^4$. To specify such a linear operator, you need $4\times 4 = 16$ numbers. If you're looking for a $2 \times 2$ matrix that represents your operator, you have to know that in general, it's impossible. Because you have $16$ degrees of freedom and $2\times 2$ matrices offer you only $4$ degrees of freedom.
Back to your question, if you want to represent $L_{A,B}$ as $L_{A,B}(X)=TX$ where $T$ is a $4 \times 4$ matrix, you need to first find a way to inject $2\times 2$ matrices into $4 \times 4$ matrices that is compatible with your representation, i.e. for any $X$, the following system should be solved: $$X = \begin{bmatrix} x_1&x_2\\ x_3&x_4\\ \end{bmatrix}$$ $$\begin{bmatrix} 9&7&6&6\\ 3&0&0&3\\ 2&0&0&2\\ 12&7&6&9\\ \end{bmatrix}\begin{bmatrix} t_1&t_2&t_3&t_4\\ t_5&t_6&t_7&t_8\\ t_9&t_{10}&t_{11}&t_{12}\\ t_{13}&t_{14}&t_{15}&t_{16}\\ \end{bmatrix}=\begin{bmatrix} 9x_1&7x_1&6x_2&6x_2\\ 3x_1&0&0&3x_2\\ 2x_3&0&0&2x_4\\ 12x_3&7x_3&6x_4&9x_4\\ \end{bmatrix}$$
However, since the first matrix is singular because its determinant is $0$ (the second and the third row are linearly dependent), injecting $2\times 2$ matrices into $4 \times 4$ matrices that is compatible with your representation fails. So, I'm afraid that your representation doesn't work.