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Let, $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & -1 \end{bmatrix}$ . Then find $ S= \{B \in M_{3\times3}(\Bbb R) :AB=BA\}$ .

My attempt:

Bare computation yielded $\{B\in M_{3\times3}(\Bbb R) : B = \begin{bmatrix} c_1 & c_2 & 0 \\ c_3 & c_4& 0 \\ 0 & 0 & c_5 \end{bmatrix} , c_1,c_2,c_3,c_4,c_5 \in \Bbb R\} \subseteq S$ but I am having trouble showing the reverse inclusion (if it is true!).

I was also tempted to use eigenvalues and eigenspaces to reach some conclusion by looking at the diagonal form of $A$ (displaying its eigenvalues), but can't get my way through!

Thanks in advance for help.

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Yes, the reverse inclusion is true. Note that if$$B=\begin{pmatrix}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{pmatrix},$$then$$AB-BA=\begin{pmatrix}0&0&2b_{13}\\0&0&2b_{23}\\-2b_{31}&-2b_{32}&0\end{pmatrix}.$$Therefore, $AB=BA$ if and only if $b_{13}=b_{23}=b_{31}=b_{32}=0$.

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  • $\begingroup$ Can you give some idea with the Eigen spaces and Eigen values. $\endgroup$ – reflexive Jun 23 '18 at 10:30
  • $\begingroup$ The block nature should make it obvious what the eigenspaces are.. $\endgroup$ – Cameron Williams Jun 23 '18 at 10:34
  • $\begingroup$ I meant some arguments by Eigen spaces and Eigen values $\endgroup$ – reflexive Jun 23 '18 at 10:46
  • $\begingroup$ @ThatIs Right now, I can't think of any. $\endgroup$ – José Carlos Santos Jun 23 '18 at 10:47
  • $\begingroup$ @JoséCarlosSantos I've tried to give an argument via Eigen spaces and all. $\endgroup$ – reflexive Oct 7 '18 at 16:31
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Opposed to the direct computation, this is an approach via eigenspaces.

Let $M,N$ be any two matrices such that $MN=NM$. Let $c$ be an eigenvalue of $M$ and let's call $E_c$ to be the eigenspace associated to the eigen value $c$ .

Then for any $v \in E_c$ , $M(Nv)=(MN)v=(NM)v=N(Mv)=N(cv)=c(Nv)$ . So we get that, $N(E_c) \subset E_c$ .

Now back to our given problem. $A$ clearly has two eigenvalues : 1 and -1 with the former having multiplicity 2 and the latter having multiplicity 1.

A routine computation shows that, $E_1 = span\{(1,0,0),(0,1,0)\}$ and $E_{-1} = span\{(0,0,1)\}$.

So we get that, $B(1,0,0) \in E_1, B(0,1,0) \in E_1, B(0,0,1) \in E_{-1} $ i.e. $\exists c_1, \dots ,c_5\in \Bbb R$ such that ,$$B(1,0,0)=c_1(1,0,0)+c_2 (0,1,0) + 0(0,0,1)$$ $$B(0,1,0)=c_3(1,0,0)+c_4 (0,1,0)+0(0,0,1)$$ $$B(0,0,1)=0(1,0,0)+0 (0,1,0)+c_5(0,0,1)$$

And hence, in the standard basis, we get, $$B=\begin{bmatrix} c_1 & c_2 & 0 \\ c_3 & c_4 & 0 \\ 0 & 0 & c_5 \end{bmatrix}$$

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