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Yes, this is a repeat, however I have not seen anyone explain it fully (or such that I can comprehend it, and believe me, I have searched thoroughly for answers).

If the (linear) endomorphisms $A,B: V \to V$ are diagonalisable, show that they are simultaneously diagonalisable $\iff AB=BA$

The initial implication is trivial. I have shown the case for when all eigenvalues are distinct. It is when there are not necessarily distinct that I cannot seem to get my head around the problem. (For instance, minimal polynomials are too unfamiliar to me to be constructive). Any links, proofs, hints or explanations are deeply, deeply appreciated.

Thanks!

2 Answers2

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Let $\lambda_1,\dots,\lambda_r$ the eigenvalues of $A$ and $E_{\lambda_i}$ the corresponding eigenspace. For any eigenvector $u\in E$, we have $$A(B(u))=B(A(u))=B(\lambda u)=\lambda B(u).$$ This proves the $E_{\lambda_i}$s are stable by $B$.

Now, since $A$ is diagonalisable, the vector space $V$ decomposes as $$V=\bigoplus_{i=1}^rE_{\lambda_i}$$ In a basis of eigenvectors for $A$ the matrix A becomes a diagonal matrix, and the matrix $B$ is a block-diagonal matrix $$\begin{pmatrix} B_1\!\\ &\!\ddots\!\\ &&\!B_r \end{pmatrix}$$ Thus it is enough to observe the restriction of (the endomorphism associated with) $B$ is diagonalisable. Take in each $E_{\lambda_i}$ a basis of eigenvectors of $B_i=B\Bigl\lvert_{E_{\lambda_i}}$. The matrix of the restriction of $A$ to this eigenspace remains diagonal, since it is $\lambda_i I_{E_{\lambda_i}}$. Finally, choose as a basis for $V$ the union of the bases of the $E_{\lambda_i}$. You obtain a basis which diagonalises simultaneously $A$ and $B$.

Bernard
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    I understand the part with the block matrix $B$ and to prove that this is diagonalisable. "Take in each $E_{\lambda_i}$ a basis of eigenvectors of $B_i$. The matrix of the restriction of $A$ to this eigenspace remains diagonal, since it is $\lambda_i I_{E_{\lambda_i } }$." I thought that the matrix of the restriction of $A$ would be like this only in the basis of eigenvectors of $A$, i.e. taking a basis of eigenvectors of $B_i$ in every $E_{\lambda_i} $ only makes it so that all blocks $B_i$ are diagonal, but not necessarily the blocks of $A$? –  Feb 28 '16 at 15:06
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    forgot to tag, thanks for your contribution btw. –  Feb 28 '16 at 15:27
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    About your objection: on each $E_{\lambda_i}$, the linear map induce by $A$ is multiplication by $\lambda_i$, by definition of an eigenspace, so its matrix in any basis is simply $\lambda_i I_{E_{\lambda_i}}$. – Bernard Feb 28 '16 at 21:40
  • Dear Sir, I had one doubt. I understand we got block diagonals of matrix B wrt eigenspace of A. and also on each restriction each block diagonal is also diagonalisation. But How it implies with using the same eigenspace of A, that I do not understand. Please Help me to understand – Curious student Mar 19 '19 at 13:07
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As stated, the claim is false. $$A=B=\begin{pmatrix}0&1\\0&0\end{pmatrix} $$ clearly commute, but are not diagonalizable.

Hagen von Eitzen
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    Yes, someone pointed out that 'mistake'. Just editted the post. Thanks for noticing. –  Feb 28 '16 at 11:55