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Say, I have two square matrices, $A$ and $B$, not necessarily Hermitian, whose eigenvalues and eigenvectors are known, and I also know if they are diagonalizable. Is there a way to figure out if their product $AB$ is diagonalizable without explicitly calculating the eigenvalues and eigenvectors of $AB$?

Rodrigo de Azevedo
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user3433489
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1 Answers1

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For example, try $A = \pmatrix{1 & 1\cr 0 & 1\cr}$ and $B = \pmatrix{1 & t\cr 0 & 1\cr}$ where $t \ne 0$. $A$ and $B$ both are non-diagonalizable with eigenvalue $1$ and eigenvector $\pmatrix{1\cr 0\cr}$. $AB = \pmatrix{1 & 1+t\cr 0 & 1\cr}$, which is also non-diagonalizable except when $t = -1$, in which case it is diagonalizable.

So you can't always tell whether $AB$ will be diagonalizable by just looking at the diagonalizability, eigenvalues and eigenvectors of $A$ and $B$ separately.

Robert Israel
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    Also by (for example) this question: https://math.stackexchange.com/questions/1084522/is-the-product-of-any-two-invertible-diagonalizable-matrices-diagonalizable, even the product of diagonalisable matrices may not be diagonalisable. – Uzai Feb 17 '23 at 13:29
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    @Uzai One might still have hoped that we can tell when the product of two diagonalizable matrices is not diagonalizable by looking at their eigenvectors. The point of the counterexample in the answer is that even those do not provide sufficient information. – Misha Lavrov Feb 17 '23 at 17:23