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Consider subspace of matrix $X \subset Mat_{n\times n}(F)$: $A_{i}X+XB_{i} = 0$, where $(A_{i},B_{i})_{i=1}^{m}$ given set of matrix with size $m$.

How can I find dimension of $X$ and their basis? I thought about consider linear combination of some special matrix from $X$ and "delete" some of them to find needed quantity.

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    I'm not sure if I understand: do you have just the two matrices $A_i,B_i$, or do you have several pairs $(A_1,B_1),(A_2,B_2),...$? – 2017-02-13
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    @Omnomnomnom I fixed it – 2017-02-13
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    The dimension will obviously depends on the matrices. For example, if $A_i = B_i = 0$ for all $i$ then your subspace is $M_n(F)$. If you pick $B_1 = 0, A_1 \in GL_n(F)$ then your subspace is $0$. So there is a big gap and I am afraid there is not much to say unless you have concrete $A_i$ and $B_i$ ... – 2017-02-13

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Let $S_i = \{X:A_iX + XB_i = 0\}$. Then your subspace is the intersection $$ S = \bigcap_{i=1}^n S_i $$ If you look up an analysis of Sylvester's equation, you can find that $S_i$ is spanned by the matrices of the form $uv^T$ such that for some $\lambda$: $u$ is an eigenvector of $A_i$ associated with $\lambda$, $v$ is an eigenvector of $B_i^T$ associated with $-\lambda$. Note that this assumes that $F$ is algebraically closed. I think that it's true whether or not $A_i,B_i$ are diagonalizable.

In particular: if you have any pair $A_i,B_i$ such that $A_i$ and $-B_i$ have no common eigenvalues, then we will have $S_i = \{0\}$ and therefore $S = \{0\}$.