I wanted to prove that If A is diagonalisable then $T\in L(V)$ given by $T(B)=AB-BA$ is diagonalisable.
My Attempt:
I know that if minimal polynomial of T has distanct linear factors then we can show that T is is diagonalisable.
As A is diagonalisable imlies it has distinct linear factor
I don't know how to give next argument .
ANy Help will be appreciated
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Arctic Char
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Curious student
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Related thread: https://math.stackexchange.com/questions/1345661/eigenvalues-of-linear-operator-fa-ab-ba. – Batominovski Sep 11 '18 at 17:04
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You can assume that $A$ is already diagonal. Then the matrix units $E_{ij}$ (the matrices all zero save for one $a$) form a basis of eigenvectors of $T$.

Angina Seng
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