Show A and B have a common eigenvalue

Question

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.

score 7 · Accepted Answer · answered Aug 28 '14 at 19:04

7

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.

answered Aug 28 '14 at 19:04

Kelenner

18,734
26
36

why does "Then there exists $U$, $V$ polynomial such that $M(X)U(X)+N(X)V(X)=1$." ?

user7610

Aug 28 '14 at 19:29

How did you infer $CN(B)V(B)=0$. ?

– user7610 Aug 28 '14 at 19:29

score 1 · Answer 2 · answered Aug 28 '14 at 18:34

1

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.

answered Aug 28 '14 at 18:34

amcalde

4,674

5

$(C v_j)$ may be 0... In which case it's not an eigenvector. How does one show there is one like that that isn't zero? – user7610 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

score 1 · Answer 3 · answered Sep 05 '14 at 06:23

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} $).

Show A and B have a common eigenvalue

3 Answers3