# 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!

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In the first problem, what are the unknowns: only $X$ or $X$ and $C$? – Davide Giraudo Mar 14 '12 at 19:54

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 $A$.

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For 1) you only show that all eigenvectors of $B$ lie in the kernel of $Y$. – WimC Mar 15 '12 at 9:41
I don't see why you say that. And it's not true: take $A=B=I$, $Y$ any matrix, $\lambda=1$, $v$ any nonzero vector. You seem to say that my argument implies that $Yv=0$? – Martin Argerami Mar 15 '12 at 13:19
$(A-\lambda)Yv = 0$ implies $Yv=0$ if $\lambda$ is not an eigenvalue of $A$. – WimC Mar 15 '12 at 13:23
Now I see what you mean. Yes, in the proof I'm not guaranteeing that $Yv\ne0$. I'm running now, but I'll think about it. – Martin Argerami Mar 15 '12 at 13:46
If you look at the Jordan form of $B$, you'll notice that the eigenvalues of $p(B)$ are $$\{p(\lambda):\ \lambda \text{ is an eigenvalue of }B\}.$$ So, if $p(\lambda)\ne0$ for every eigenvalue of $B$, then $0$ is not an eigenvalue of $p(B)$ and this means that $p(B)$ is invertible. For you second question, yes. – Martin Argerami Mar 23 '12 at 17:50