# Find matrices that commute with $\operatorname{Diag}(1,1,-1)$.

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!

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 $b_{13}=b_{23}=b_{31}=b_{32}=0$.

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

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, $$B=\begin{bmatrix} c_1 & c_2 & 0 \\ c_3 & c_4 & 0 \\ 0 & 0 & c_5 \end{bmatrix}$$