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Given a matrix $A \in K^{n \times n}$, which is diagonalisable, I want to prove that the subspace, containing all $n \times n$ matrices, which commute with $A$, has at least dimension $n$.

I have already proven that if two matrices share a basis of common eigenvectors, then they commute, but I do not think this is something I can use here. I am trying to find $n$ linearly independent matrices, which commute with $A$ and I know that I can diagonalise $A$ to a diagonal matrix, which has eigenvalues as it's entries. I do not quite know, however, how to prove the statement above. Any help would be greatly appreciated.

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    $\begingroup$ More generally, if $A$ is diagonalisable, with distinct eigenvalues $\lambda_1, \ldots, \lambda_r$ and respective multiplicities $m_1,\ldots,m_r$, the space of commuting matrices has dimension exactly $\sum_{i=1}^r m_i^2\geq \sum_{i=1}^r m_i = n$ $\endgroup$ – Gabriel Romon Mar 14 '17 at 21:28
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Given that you can write $$A= T \Lambda T^{-1}$$ where $\Lambda$ is diagonal. Take a look at the $n$-dimensional space of matrices $$ B= T D T^{-1}$$ with $D$ an arbitrary diagonal matrix.

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  • $\begingroup$ Just to be sure, that space is $n$ - dimensional, since I can choose the diagonal matrix $D$ accordingly, so I only end up with the canonical basis matrix? $\endgroup$ – AxiomaticApproach Mar 14 '17 at 21:40
  • $\begingroup$ @M3xr: I am not sure I understand your question. The relevant fact is that the diagonal matrices form an $n$-dimensional vector space. (just take the individual elements along the diagonal as basis) $\endgroup$ – Fabian Mar 14 '17 at 21:41
  • $\begingroup$ I see, you are totally right. Thank you $\endgroup$ – AxiomaticApproach Mar 14 '17 at 21:44

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