Good properties of weakly commuting matrices? Let $A$ and $B$ be $n\times n$ matrices over $\mathbb{C}$. If $AB=BA$, we know that we can simultaneously diagonalize $A$ and $B$ (or make them Jordan canonical form). What if they are weakly commutative in the sense that $AB=cBA$ for some $c\in \mathbb{C}^{\times}$? What can we say about $A$ and $B$? 
I am sorry being a bit vague, but I came across with this kind of matrices in my little project and wonder what we can say about them. 
 A: I don't think you can say much. Look at the following example:
Let $\lambda\in\mathbb C$ be an $n^{\rm th}$ root of unity (say $\lambda=e^{2\pi i/n}$). Put
$$
A=\begin{bmatrix}1\\ &\lambda \\ & & \ddots \\ & & &\lambda^{n-1}\end{bmatrix},\ 
B=\begin{bmatrix} 0& 1& 0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots \\1&0&0&\cdots&0\end{bmatrix}
$$
Then both $A,B$ have eigenvalues $1,\lambda,\lambda^2,\ldots,\lambda^{n-1}$. Both are normal (they are unitaries) so they are diagonalizable, and $AB=\lambda BA$. 
But $e_1,\ldots,e_n$ (the canonical basis) is a basis of eigenvectors of $A$, while 
$$
\{(1,\alpha,\alpha^2,\ldots,\alpha^{n-1}):\ \alpha=\lambda^k,\ k=0,\ldots,n-1\}
$$ is a basis of eigenvectors for  $B$. As the vectors of either basis fails to be eigenvectors for the other matrix, we conclude that $A,B$ are not simultaneously diagonalizable. 
Note also that the condition $AB=cBA$, for $c\ne1$, places a strong restriction on what the eigenvalues of $AB$ are. Because the eigenvalues of $AB$ are the same as those in $BA$; but here $AB=cBA$, so we conclude that $BA$ and $cBA$ have the same eigenvalues. This in turn means that the eigenvalues of $BA$ are of the form $c^k\alpha$, $k=0,\ldots,n-1$ and some $\alpha\in\mathbb C$. This forces things like if any of $A,B$ is singular, then $BA$ (and thus $AB$) will have all eigenvalues equal to zero.
