# Show A and B have a common eigenvalue

Let A, B and C complex square matrices such that: $C\neq 0$ and $AC=CB$ prove that A and B has a common eigenvalue.

It's worth mentioning that earlier in the assignment I have proved that $A^{n}C=CB^{n}$, but I'm not sure how to use it.

This is taken from a linear algebra 2 course.

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Let $M$ the minimal polynomial of $A$, and $N$ the minimal polynomial of $B$. We are going to show that $M$ and $N$ have a commun root (and this will prove the assertion). Suppose not. Then there exists $U$, $V$ polynomial such that $M(X)U(X)+N(X)V(X)=1$. Now you have shown that $A^nC=CB^n$ for all $n$. This imply $M(A)C=CM(B)=0$. Hence $CM(B)U(B)=0$ and of course $CN(B)V(B)=0$. We get $C(M(B)U(B)+N(B)V(B))=0=CI=C$, a contradiction.

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1. why does "Then there exists $U$, $V$ polynomial such that $M(X)U(X)+N(X)V(X)=1$." ? – fish.frog Aug 28 '14 at 19:29
2. How did you infer $CN(B)V(B)=0$. ? – fish.frog Aug 28 '14 at 19:29
@fish.frog 1) $M(X)$ and $N(X)$ have no commun root, so their GCD is $1$, and I apply Bezout. 2) As $N$ is the minimal polynomial of $B$, we have $N(B)=0$. – Kelenner Aug 28 '14 at 19:31

Let $v_j$ be an eigenvector of $B$ with eigenvalue $\lambda_j$.

Then

$A C v_j = C B v_j \Rightarrow A (C v_j) = C (\lambda_j v_j) \Rightarrow A (C v_j) = \lambda_j (C v_j)$

Therefore $(C v_j)$ is an eigenvector of $A$ with the same eigenvalue.

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$(C v_j)$ may be 0... In which case it's not an eigenvector. How does one show there is one like that that isn't zero? – fish.frog Aug 28 '14 at 18:36
@fish.frog, what does it mean $C\ne 0$ – Alexander Vigodner Aug 28 '14 at 18:58
@AlexanderVigodner The matrix $C$ is nonzero, but the product of $C$ with that vector could yield a zero vector. – imranfat Aug 28 '14 at 19:00
If $Cv \neq 0$ for all $v$ then there would be a much simpler proof. – copper.hat Aug 28 '14 at 19:27

It has been bugging me to find an answer without (directly) using the minimal polynomial. Here it is:

Since $A^nC=CB^n$, it is easy to see that for any $\lambda$ and $k\ge 0$ we have $(A-\lambda I)^k C = C (B-\lambda I)^k$.

Suppose $\lambda \not \in \sigma(A)$. Then $C=(A-\lambda I)^{-k} C (B-\lambda I)^k$. In particular, if $v \in \ker (B-\lambda I)^k$, then $Cv=0$.

Hence if the spectra do not overlap, we must have $C=0$ (since $\mathbb{C}^n ={+} _{\lambda \in \sigma(B)} \ker (B-\lambda I)^{m_\lambda}$).

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