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More specifically: if $A$ & $B$ are diagonalizeable, than is it correct to say that $AB$ is diagonalizeable? (Hints would be more appreciated)

asaf92
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  • Not true in general. See also http://math.stackexchange.com/questions/1215569/under-what-conditions-is-the-product-of-two-invertible-diagonalizable-matrices-d. – parsiad Apr 25 '16 at 13:18
  • And the answer to the question in the title is also "no". Consider the case where $A=B=D$ is the 2x2 nilpotent Jordan block and $E=A^T$. – user1551 Apr 25 '16 at 13:22

2 Answers2

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Hint: Try to see what happens in a concrete example where $A,B$ are diagonalizable but $AB \neq BA$ (so $A$ and $B$ don't commute).

levap
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  • In fact, any matrix over the complex numbers can be written as a product of two diagonalizable matrices (see Polar Decomposition of matrices) but not all matrices are diagonalizable. – levap Apr 25 '16 at 13:23
  • I tried an example: $$A = \begin{pmatrix} 2 & 2 \ 1 & 3\end{pmatrix}$$ $$B = \begin{pmatrix} 1 & 1 \ 0 & 2\end{pmatrix}$$ $AB\not = BA$ but $AB$ is diagonalizeable. – asaf92 Apr 26 '16 at 12:21
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we have: $$ A=P^{-1}DP \quad \land \quad B=Q^{-1}EQ \quad \Rightarrow \quad AB=P^{-1}DPQ^{-1}EQ $$ So we can say that if $P=Q$ ( i.e. $A$ and $B$ are simultaneously diagonalizable) than $AB$ is diagonalizable.

Emilio Novati
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