Two simple eigenvector problems I am new to the notion of eigenvector. In my book right after the definition there are the following problems.
1) If $A\in M_{n\times n}(\mathbb{C})$ and $B\in M_{m\times m}(\mathbb{C})$ don't have common eigenvalues, then matrix equation $AX-XB=C$, where $X\in M_{n\times m}(\mathbb{C}), C\in M_{n\times m}(\mathbb{C})$, has exactly one solution.
2) If $A\in M_{n\times n}(\mathbb{C}), B\in M_{n\times n}(\mathbb{C})$ satisfy $AB=BA$, then they have common eigenvector.
I am completely stuck, as I don't know how to approach such problems. Thanks!
 A: For 1), if $X_1,X_2$ are two solutions, we put $Y=X_1-X_2$ and we get
$$
0=AY-YB,
$$
so that $AY=YB$. Then we have
$$
A^2Y=A(AY)=A(YB)=(AY)B=YB^2.
$$
By induction we deduce that $A^kY=YB^k$ for all $k\in\mathbb{N}$, and by linearity we get
$$
p(A)Y=Yp(B),\ \ p\in \mathbb{C}[x].
$$
Now, since $A$ and $B$ have no common eigenvalues, we can choose a polynomial $p$ such that $p(A)=0$, $p(B)$ invertible (this would be a polynomial that is zero on all eigenvalues of $A$, and is nonzero on all eigenvalues of $B$). But then we get
$$
0=Yp(B)
$$ 
with $p(B)$ invertible, so $Y=0$, i.e. $X_1=X_2$. That is, the solution is unique.
We now consider the map $X\mapsto AX-XB$ from $M_{n\times m}(\mathbb{C})$ into itself; this map is linear. By the uniqueness above, it is also injective; but in a finite-dimensional environment, this implies that it is also surjective. So given any $C$, there exists $X$ such that $AX-XB=C$. Thus there always exists a solution, and by above it is unique.
For 2), let $\lambda$ be an eigenvalue of $A$, and let
$$
M=\{v:\ Av=\lambda v\}.
$$
For $v\in M$, $A(Bv)=BAv=\lambda Bv$, so $BM\subset M$. So we can consider $B$ as an operator restricted to $M$; there has to be an eigenvalue $\mu$ of $B|_M$, and so there exists $w$ such that $Bw=\mu w$. So $w$ is an eigenvector for $B$, and since it is in $M$ it is also an eigenvector for $A$. 
