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Let $A$ be a $n \times n$ non-singular matrix having distinct eigenvalues. If $B$ is a matrix satisfying $A B = B A^{-1}$, show that $B^2$ is diagonalisable.

Answer:

Let $\lambda_i, \ i=1,2,3, \cdots, n$ be the $n$ distinct eigenvalues.

Now, $AB=BA^{-1} \Rightarrow B=ABA $

But then how to proceed?

Rodrigo de Azevedo
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MAS
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1 Answers1

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We can show that $AB^2=ABABA=B^2A$. So $B^2$ and $A$ commute. Then the claim follows from these duplicates (replacing $B$ by $B^2$):

$AB=BA$. Prove $B$ is diagonalizable.

If $AB=BA$, show that $B$ is diagonalizable.

Indeed, we have $AB^2=(AB)B=ABABA$ and $B^2A=B(BA)=ABABA$.

Dietrich Burde
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