While studying linear algebra, I saw an example of a pair of $2 \times 2$ or $n \times n$ diagonalizable matrices, the product of which is not diagonalizable. Is there a similar example when I replace the condition "diagonalizable" by "invertible and diagonalizable"?
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A counterexample:$$\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix}=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ The result is a well known nondiagonalizable matrix, the left matrix of the product is diagonal already, and the right matrix can be written as $$\begin{pmatrix}1&1\\0&-1\end{pmatrix}=\begin{pmatrix}-1&1\\2&0\end{pmatrix}\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}-1&1\\2&0\end{pmatrix}^{-1}.$$
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1As painfully simple as that. +! – Timbuc Dec 29 '14 at 12:57
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There's a tiny typo in the last line -- the inner matrix should be diagonal. – Clement C. Dec 29 '14 at 12:57