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Let $V$ be a finitely generated inner product space above $\mathbb{F}$. Denote by ${\rm End}\left(V\right)$ the space linear transformations $V\to V$. For $A\subseteq{\rm End}\left(V\right)$ Denote $$C\left(A\right)=\left\{ X\in{\rm End}\left(V\right)|X\circ S=S\circ X\quad\forall S\in A\right\} $$

For $S\in{\rm End}\left(V\right)$, denote $C\left(S\right)=C\left(\left\{S\right\} \right)$ and denote

$${\rm Pol}\left(S\right)=\left\{ p\left(S\right):p\in\mathbb{F}\left[t\right]\right\} $$

Show that if $S$ is diagonalizable then ${\rm Pol}\left(S\right)=C\left(C\left(S\right)\right)$.

I know that is $S$ has distinct eigenvalue then any transformation in $C(S)$ must be a polynomial in $S$, but apart from that I'm not really sure how to proceed.

One side I was able to do:

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Let $p\in\mathbb{F}\left[t\right]$, then $p\left(S\right)\in{\rm Pol}\left(S\right)$ and we need to prove $p\left(S\right)\in C\left(C\left(S\right)\right)$, that is, for all $T\in C\left(S\right)$ we want to prove that $p\left(S\right)\circ T=T\circ p\left(S\right)$. Let $T\in C\left(S\right)$. Then by definition $T\circ S=S\circ T$. Then $$p\left(S\right)\circ T=\left(\sum_{i=0}^{n}a_{i}S^{i}\right)\circ T=\sum_{i=0}^{n}a_{i}S^{i}T=\sum_{i=0}^{n}a_{i}TS^{i}=T\sum_{i=0}^{n}a_{i}S^{i}$$ as required.

A bit stuck about the other side.

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    $\begingroup$ Possible duplicate of Commuting linear maps $\endgroup$ – Dietrich Burde Dec 28 '15 at 18:58
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/65012/… $\endgroup$ – Dietrich Burde Dec 28 '15 at 19:00
  • $\begingroup$ @DietrichBurde This is not quite a duplicate of that. There is no assumption here that the minimal and characteristic polynomials are the same. $\endgroup$ – Omnomnomnom Dec 28 '15 at 19:02
  • $\begingroup$ By the way: the term is an inner product space over $\Bbb F$, not above. $\endgroup$ – Omnomnomnom Dec 28 '15 at 19:12
  • $\begingroup$ $C(C(S)) = Pol(S)$ is true for all $S$. No need to assume $S$ diagonalizable. $\endgroup$ – Orest Bucicovschi Dec 28 '15 at 21:33
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Sketch of proof: With an appropriate choice of basis, we can take $S$ to be a diagonal matrix. That is, $$ S = \pmatrix{\lambda_1 I_{m_1} \\ & \lambda_2 I_{m_2}\\ &&\ddots\\ &&& \lambda_k I_{m_k} } $$ here, $\lambda_i$ are the (distinct!) eigenvalues of $S$, $I_m$ is the $m \times m$ identity, and $m_{i}$ is the multiplicity of the $i$th eigenvalue.

First, find that $C(S)$ is the set of matrices of the form $$ \pmatrix{A_1\\&A_2\\&&\ddots \\&&&A_k} $$ where $A_i$ is an arbitrary $m_i \times m_i$ matrix.

Next, find that $C(C(S))$ is the set of matrices of the form $$ \pmatrix{\mu_1 I_{m_1} \\ & \mu_2 I_{m_2}\\ &&\ddots\\ &&& \mu_k I_{m_k} } $$ where $\mu_i \in \Bbb F$ are arbitrary. It should be clear from this point that $Pol(S) \subset C(C(S))$ (though for you, this is unnecesssary). For the reverse inclusion, it suffices to note that for any $\mu_1,\mu_2,\dots,\mu_k$, it suffices to select an interpolating polynomial satisfying $$ p(\lambda_i) = \mu_i \qquad i=1,\dots,k $$

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    $\begingroup$ The key to going through the proof this way is a certain degree of comfort with block-matrix multiplication. $\endgroup$ – Omnomnomnom Dec 28 '15 at 19:03
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    $\begingroup$ I keep putting discussion of the general case (not quite the style of this question) at math.stackexchange.com/questions/92480/… $\endgroup$ – Will Jagy Dec 28 '15 at 20:20
  • $\begingroup$ @WillJagy very nice. Feel free to add this answer to the collection if you find it sufficiently relevant. $\endgroup$ – Omnomnomnom Dec 28 '15 at 20:26
  • $\begingroup$ That makes sense.. Had troubles using the fact S is diagonalizable, but this makes it doable for me. $\endgroup$ – Nescio Dec 29 '15 at 8:44

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