Is the Product of Two Diagonalizable Matrices Always Diagonalizable?

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SUMMARY

The product of two diagonalizable matrices A and B is not always diagonalizable, even if both matrices are invertible and of the same order. A counterexample is provided with matrices A = [1, 1; 2, -1] and B = [1, 2; 2, 1], where their product AB = [3, 3; 0, 3] is not diagonalizable. However, if A and B commute (i.e., AB = BA), then their product is diagonalizable, as they can be simultaneously diagonalized using the same invertible matrix S.

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Yankel
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Hello

I have a little question

if A and B are both invertible and diagonalizable matrices (from the same order), is A*B a diagonalizable matrix ? why ?

I have not got a clue...

thanks !
 
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Yankel said:
if A and B are both invertible and diagonalizable matrices (from the same order), is A*B a diagonalizable matrix ? why ?

It is not true. Verify that $A=\begin{bmatrix}1&{\;\;1}\\{2}&{-1}\end{bmatrix}$ and $ B=\begin{bmatrix}{1}&{2}\\{2}&{1}\end{bmatrix}$ are invertible and diagonalzable matrices on $\mathbb{R}$, however $AB=\begin{bmatrix}{3}&{3}\\{0}&{3}\end{bmatrix}$ it is not diagonalizable.
 
Yankel said:
Hello

I have a little question

if A and B are both invertible and diagonalizable matrices (from the same order), is A*B a diagonalizable matrix ? why ?

I have not got a clue...

thanks !

Not always. But when A,B are interchangable, i.e., AB=BA, then AB IS diagonalizable since then A and B are simultaneously diagonalizable, i.e., we can find the same invertible matrix
S such that $A=S^{-1} D_1 S, B=S^{-1} D_2 S$, where $D_1,D_2$ are both diagonal matrices. Thus we have $AB=S^{-1}D S$ with $D=D_1D_2$ a diagonal matrix.
 

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