# Product of an invertible diagonal matrix and a diagonalizable matrix is diagonalizable?

I encountered one problem. Suppose $$\textbf{A}$$ is a diagonal invertible matrix and $$\textbf{B}$$ is a diagonalizable matrix of same size. Is the product matrix $$\textbf{AB}$$ is diagonalizable?

Here is how I proceeded.

All we want is to find an invertible matrix $$\textbf{Q}$$ such that $$\textbf{AB}=\mathbf{Q}(\mathbf{D}) \mathbf{Q}^{-1},$$ for some diagonal matrix $$\mathbf{D}$$. (This $$\mathbf{D}$$ is actually called as a similar matrix to that of $$\mathbf{AB}.)$$

I begin from the information about $$\textbf{B},$$ viz. $$\mathbf{B}$$ is diagonalizable. Then by definition of diagonalizability, there exists some invertible matrix $$\mathbf{P}$$ satisfying $$\mathcal{D}=\mathbf{P}^{-1}\mathbf{B}\mathbf{P},$$ for some diagonal matrix $$\mathcal{D}.$$ This is same as $$\mathbf{B}=\mathbf{P}\mathcal{D}\mathbf{P}^{-1}.$$ Let us premultiply this last equation with $$\mathbf{A},$$ implies $$\mathbf{AB}=\mathbf{A}(\mathbf{P}\mathcal{D}\mathbf{P}^{-1}).$$

Now what kind of techniques we suppose to use? Any help is appreciated.

Let $$A=\begin{bmatrix} \frac12 & 0 \\ 0 & 1\end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 2 \\ 0 & 1\end{bmatrix}$$.
The eigenvalues of $$B$$ are $$1$$ and $$2$$, hence it is diagonalizable. However,
$$AB = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$$ which is not diagonalizable.
• ach, ninjed  – Exodd Apr 11 at 9:16