If A and B are diagonalizable then so is AB When we have to n×n matrices that can be made diagonal (maybe not in the same basis), is it true that the same works for their product?
 A: That's not true in general.
let $A=\begin{pmatrix}
1&1\\0&\frac12
\end{pmatrix}$, $B=\begin{pmatrix}1&1\\0&2 \end{pmatrix}$, those 2 matrices are clearly diagonalizable since they have distinct eigenvalues, while $AB=\begin{pmatrix}1&3\\0&1\end{pmatrix}$ isn't.
A: No, It's not true in general. Let $A=S_1\Lambda_1S_1^{-1}$ and and $B=S_2\Lambda_2 S_2^{-1}$, Then $AB=S_1\Lambda_1S_1^{-1}S_2\Lambda_2 S_2^{-1}$. So if $S_1=S_2$ then $AB=S_1\Lambda_1S_1^{-1}S_1\Lambda_2 S_1^{-1}$$=S_1\Lambda_1\Lambda_2 S_1^{-1}$ and we know that product of two diagonal matrix is a diagonal matrix. 
Next the question is when, $S_1=S_2$
In G. Strang's Linear Algebra and its Applications it is given that, if $A$ and $B$ are diagonalizable matrices of the form  such that $AB=BA$, then their eigenvector matrices $S_1$ and $S_2$ (such that $A=S_1\Lambda_1S_1^{-1}$ and $B=S_2\Lambda_2 S_2^{-1}$) can be chosen to be equal: $S_1=S_2$.
You can find detailed proof in the following link:
$AB=BA$ with same eigenvector matrix
