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The space of matrices that commute with a given matrix $A\in \mathbb{C}^{n\times n}$ is a subspace of the vector space of all matrices $\mathbb{C}^{n\times n}$.

There must exist a projection operator upon this subspace, some $P_A$ such that $$\forall M \in \mathbb{C}^{n\times n}: [P_A M,A]=0$$ Question: Is there some useful expression for $P_A$ in terms of $A$?

Context

I'm looking for a way to interpolate matrices without breaking commutation. For example, I may want to construct $$f:[0,1]\rightarrow \mathbb{C}^{n\times n}$$ $$g:[0,1]\rightarrow \mathbb{C}^{n\times n}$$ knowing $f(0),f(1),g(0),g(1)$ and $[f(0),g(0)]=[f(1),g(1)]=0$, in such at way that $[f(x),g(x)]=0$ remains true for all $x$.

Wouter
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  • There are infinitely many projections onto a given subspace. Here, you could choose the orthogonal projection with respect to the Hilbert-Schmidt inner product. – amsmath Sep 03 '18 at 17:03
  • For more information on that subspace you may want to consult https://math.stackexchange.com/questions/1379878/cm-a-in-m-n-mathbbc-mid-am-ma-is-a-subspace-of-dimension-at-least – amsmath Sep 03 '18 at 17:25

1 Answers1

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Consider the case where $A$ is diagonalizable with $n$ distinct eigenvalues: $A = S \Lambda S^{-1}$ where $\Lambda$ is diagonal. Then it's easy to see that the matrices that commute with $A$ are all those that are diagonalized using the same $S$: $\{M: \;S^{-1} M S \text{ is diagonal}\}$. A projection on this space is $P_A(M) = S P_D(S^{-1} M S) S^{-1}$ where $P_D$ makes all off-diagonal elements $0$ and leaves the diagonal elements unchanged: $$P_D(X)_{ij} = \cases{X_{ii} & if $i=j$\cr 0 & otherwise}$$

Robert Israel
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  • Robert, this is really nice. I just checked that this projection does not depend on the choice of $S$ and is the orthogonal projection onto the subspace with respect to the inner product $\langle X,Y\rangle = \operatorname{Tr}((S^{-1}XS)(S^{-1}YS)^*)$. – amsmath Sep 03 '18 at 19:33