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?


Question: Where exactly is the London Shoreditch National Express Coach Stop?
Answer: The National Express website is presumably the one you mean, where it shows it off the African coast.

However, if you look further down the page, it says:


  London (Shoreditch)
  
  Bethnal Green Rd (to Stansted Airport: Stop J
  
  opp Overground Stn; or
  
  from Stansted: Stop K)


Bethnal Green road on Google Maps clearly shows the road  running west-east, with the Overground station indicated on the same map.

Hope that helps!