Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The proof given by my textbook is highly non-satisfying. The author adopted some magic-like "reductio ad absurdum" and the proof (although is correct) didn't reveal the nature of this problem. I made my own effort into it and tried a different approach. Yet I can't finish it.

Let $\mathscr{F}$ be a commuting family in $M_n(\mathbb{C}^n)$, and $A\in\mathscr{F}$, then $A$ has $n$ eigenvalues. We pick one, say $\lambda$. Let $x$ be one of its eigenvector.

We can easily prove that, if $A$ has no other eigenvector with eigenvalue $\lambda$ that linearly independent with $x$, which means that $\{cx|c\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$, then $x$ is a common eigenvector. Because $\forall B\in\mathscr{F}$ and $\forall y\in \{cx|c\in\mathbb{C}\}$, $$A(Bx)=ABx=BAx=B(Ax)=B(\lambda x)=\lambda (Bx)$$, so that $Bx$ has to be in $\{cx|c\in\mathbb{C}\}$, that is, $Bx=c_0x$ for some $c_0\in\mathbb{C}$, which means $x$ is a eigenvector of $B$ too.

But what if there are vectors satisfying $Ax=\lambda x$ that's not in $\{cx|c\in\mathbb{C}\}$? Well, then we should have a set of linearly independent eigenvectors $\{x_1,x_2,...,x_k\}$, that $\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$.

Now, I have a reasonable hypothesis that there exists some $x=c_1x_1+c_2x_2+...+c_kx_k$, that can be proven to be a common eigenvector of $\mathscr{F}$. I've tried some approaches to prove it but all failed.

Do you guys believe it's true? And if it is true then how do I prove it?

share|improve this question
2  
Dear Voldemort, This answer is relevant. Regards, –  Matt E Sep 16 '12 at 3:34

1 Answer 1

up vote 2 down vote accepted

Eureka! I figured it out! It's quite silly because in fact I was so close.

So yes, my hypothesis is true.

Actually, Since $W_1=\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}$ is obviously a subspace of $W_0=\mathbb{C}^n$, and we can easily deduce (same approach as the first part of my proof) that $Bx\in W_1$ as long as $x\in W_1$, we obtain that $B$ is $W_1$-invariant, or we can say $B$ is also a linear transformation on $W_1$, so that there are $k$ eigenvalues of $B$ in $W_1$.

Now, pick one of them, say $\lambda_B$. As we can clearly see it's the first part of my proof all over again. For any $C\in\mathscr{F}$, we can get a new subspace $W_2\subseteq W_1\subseteq W_0$, that $C$ is $W_2$-invariant, and go ahead we get $W_3\supseteq W_4\supseteq ...$

Note that, $W_0$ has a finite dimension $n$, and every time we have $1\le\dim{W_{i+1}}\le\dim{W_i}$ so there gotta be a subspace $$W_\infty=\bigcap_i{W_i}$$ with $\dim{W_\infty}\ge1$, that $\forall x\in W_\infty(x\ne 0)$ is an common eigenvector of all the matrices in $\mathscr{F}$.


As inspired by Matt.E's answer in that question, we know that $\forall A\in\mathscr{F}$, if $A$ has $k$ eigenvalues, then there are at least $k$ common eigenvectors in $\mathscr{F}$. Simply because we can do the same process for each of its eigenvalues.

share|improve this answer
1  
What if the family of commuting matrices is uncountable? –  Dimitrios Ntalampekos Aug 28 '13 at 6:06

Your Answer

 
discard

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.