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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.

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Answer is negative. It need not be diagonalizable.

Example:

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.

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    $\begingroup$ ach, ninjed $ $ $\endgroup$ – Exodd Apr 11 at 9:16

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