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?


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.

  • 2
    $\begingroup$ What if the family of commuting matrices is uncountable? $\endgroup$ – Dimitris Aug 28 '13 at 6:06

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