If $\bigoplus\limits_{i=1}^{k} Eig(A,\lambda_i)=\bigoplus\limits_{i=1}^{k} Eig(B,\mu_i)$ then $A$ and $B$ commute

Asked Mar 10 '16 at 16:07

Active Mar 10 '16 at 18:08

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Let $A,B$ be two $n\times n$-matrices over $\mathbb K$ and assume $\bigoplus\limits_{i=1}^{k} Eig(A,\lambda_i)=\bigoplus\limits_{i=1}^{k} Eig(B,\mu_i)=\mathbb K^n$, where $\lambda_i, \mu_i$ are distinct eigenvalues of $A$ and $B$ respectively, show that if $\forall i, \exists j$ s.t. $Eig(B,\mu_i)=Eig(A,\lambda_j)$ then $A$ and $B$ commute

Does it follow from $\mathbb K$ and assume $\bigoplus\limits_{i=1}^{k} Eig(A,\lambda_i)=\bigoplus\limits_{i=1}^{k} Eig(B,\mu_i)=\mathbb K^n$ that $A$ and $B$ are diagonalizable, or more than that ?

the question looks like the converse of this

edited Apr 13 '17 at 12:21

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asked Mar 10 '16 at 16:07

inequal

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We can say that $A$ and $B$ are diagonal matrices with respect to the basis of eigenvectors – Ben Grossmann Mar 10 '16 at 16:19
@Omnomnomnom so they're simultaneously diabonalizable – inequal Mar 10 '16 at 18:48
precisely ${}{}{}{}{}{}{}$ – Ben Grossmann Mar 10 '16 at 19:01

If $\bigoplus\limits_{i=1}^{k} Eig(A,\lambda_i)=\bigoplus\limits_{i=1}^{k} Eig(B,\mu_i)$ then $A$ and $B$ commute

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