Eigenvectors problem So I have a question which I do not know how to solve.
Let $A,B \in M_{n}(\mathbb{C})$ and let $AB=BA$.
I have to prove that there exists a common eigenvector for both $A$ and $B$.
How? I have no clue.
And by the way, I think the next statement is true, but if not please notify me.
$\forall X,Y \in M_{n}(\mathbb{C})$  the characteristic polynomials of $XY$ and $YX$ are the same, even if $XY \neq YX$.
( "A matrix $A$ over a field containing all of the eigenvalues of $A$ is similar to a triangular matrix." )
Thanks in advance!
 A: If $AB=BA$ and $Av=\lambda v$, then $ABv=BAv=\lambda Bv$, so $B$ maps eigen-spaces of $A$ into themselves. Hence we can restrict the map $B$ on an eigen-space $E$. Then this restricted mapping $B\mid_E$ must have some eigen-vector in $E$, because we are working over $\mathbb C$. So there is one common eigen-vector for both matrices.
For your second statement, see this question for a solution. Notice that in our case we can avail of the LU and UL decompositions of matrices of the form $M_1=\begin{pmatrix}tI&0\\*&tI-YX\end{pmatrix}, M_2=\begin{pmatrix}tI-XY&*\\0&tI\end{pmatrix}$ as done by julien in the linked answer, to deduce that characteristic polynomials of $XY$ and of $YX$ are equal.
Hope this helps.  
A: I have googled a possible solution.
If $dimN(A-\lambda I)=k$, find a basis of $N(A-\lambda I)$ as $\{x_1,x_2,...,x_k\}$, and make it as $X$, then $(A-\lambda I)X=O$.
We can see that $$(A-\lambda I)BX=ABX-\lambda BX=B(A-\lambda I)X=O.$$
So, $BX=XC$, that is, $BX$ can be represented by $\{x_1,x_2,...,x_k\}$.
Let $Cz=\beta z$, then $$BXz=XCz=\beta Xz.$$
So, $Xz$ is an eigenvector of $B$.
And $Xz$ can be represented by $\{x_1,x_2,...,x_k\}$, then $(A-\lambda I)Xz = O$.
So, $Xz$ is the common eigenvector of $A$ and $B$.
A: Considering diagonalizable matrices, this might probably work:
Consider $\mathbf{B=\mathbf{V^H\Lambda}V}$. Since $\mathbf{AB=BA}$, $\mathbf{AV^H\Lambda V=V^H\Lambda VA}$, meaning that $\mathbf{(VAV^H)\Lambda=\Lambda(VAV^H)}$. 
Denoting $\mathbf{VAV^H}$ by $\mathbf{C}$ we see that $\mathbf{C}$ commutes with a diagonal matrix $\mathbf{C\Lambda=\Lambda C}$.
Expanding this last equation shows that $\mathbf{C}$ itself is diagonal.
I hope this is right. Please lemme know.
A: Also, wrt to the last part of your question, the two matrices $\mathbf{AB}$ and $\mathbf{BA}$ have the same eigenvalues, up to zero eigenvalues (the larger of the two will have all the eigenvalues of the smaller plus zero eigenvalues), right?
I mean: $$\mathbf{ABv}=\lambda\mathbf{v}$$ 
$$\Rightarrow \mathbf{BA(Bv)}=\lambda\mathbf{(Bv)}.$$
So, doesn't this mean they have the same eigenvalues? This means that the characteristic polynomial of one is $\lambda^k$ times the polynomial of the other.
Dunno if I misunderstood this part of your question.
A: It follows from the following statement, which is easy to verify:
If $S,T:V\to V$ are two linear transformations, and $ST=TS$, then $\ker S$ is $T$-invariant. 
