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!

Where exactly is the London Shoreditch National Express Coach Stop?
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?
