Eigenvectors problem

Question

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!

Answer 1

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.

Answer 2

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 $B$.

Answer 3

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.

Answer 4

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.

Answer 5

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.