Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
In the first problem, what are the unknowns: only $X$ or $X$ and $C$? – Davide Giraudo Mar 14 '12 at 19:54
up vote 3 down vote accepted

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 $A$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.