2

I was wondering, if $D$ and $A$ are similar matrices, over $\mathbb{R,C}$, that is $D=S^{-1}AS$ and $DC=CD$, for some $C$, must $A$ commute with $C$?

For some reason, this one is slipping from me. I want to say yes, and that it is a common fact, but I cannot seem to remember it. If so, could someone leave an example/explanation?

9301293
  • 2,264
  • 1
    The answer is no, it isn't true. – Sam Clearman May 13 '15 at 19:57
  • 1
    @SamClearman, I suspect the OP is guessing at possible methods for solving http://math.stackexchange.com/questions/1279949/if-a-matrix-commutes-with-two-others-must-the-other-two-commute from about twelve hours ago – Will Jagy May 13 '15 at 20:12
  • @WillJagy Yes, OP is. – 9301293 May 14 '15 at 02:02
  • phatty, take another look at my answer and the comments that follow it, and please do the exercises, to see what conclusions one can draw about a matrix that commutes with a diagonal matrix with all different diagonal entries. – Will Jagy May 14 '15 at 02:19

1 Answers1

4

Don't see why. However, let $C = S^{-1} F S,$ which we can do by defining $F = S C S^{-1}.$ Your $DC=CD$ becomes $$ S^{-1} A S S^{-1} F S = S^{-1} F S S^{-1} A S, $$ $$ S^{-1} A F S = S^{-1} F A S, $$ $$ A F = F A. $$

Your problem of about eleven hours ago deals with a diagonalizable matrix where the resulting diagonal has $n$ distinct entries on the diagonal, matrices, $n$ by $n.$ Please do these calculations: $$ \left( \begin{array}{rr} 5 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{rr} p & q \\ r & s \end{array} \right) = ? $$

$$ \left( \begin{array}{rr} p & q \\ r & s \end{array} \right) \left( \begin{array}{rr} 5 & 0 \\ 0 & 7 \end{array} \right) = ?? $$

So: every matrix commutes with the identity matrix. BUT, what kind of matrices commute with a diagonal matrix that has all diagonal elements different?

If the first pair was not enough, do $$ \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 11 \end{array} \right) \left( \begin{array}{rrr} r & s & t \\ u & v & w \\ x & y & z \end{array} \right) = ? $$

$$ \left( \begin{array}{rrr} r & s & t \\ u & v & w \\ x & y & z \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 11 \end{array} \right) = ?? $$

Will Jagy
  • 139,541
  • Generalized eigenspaces must be the same for commutativity. A diagonal matrix has all it's generalized eigenvalues along the diagonal, so all of the generalized eigenspaces' dimensions must be 1. I.e. it must be diagonalizable. – mathreadler May 13 '15 at 20:36
  • @mathreadler, it is more specific than that. Try my little calculation exercises. – Will Jagy May 13 '15 at 20:40
  • Yes they must have the same $S$ matrices in the jordan decomposition. I.e. must also be diagonal. – mathreadler May 13 '15 at 20:47
  • @mathreadler, yes, and you don't need any apparatus to prove that, just the exercises above interpreted. Meanwhile, I collected a bunch of equivalent conditions about commuting matrices at http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 which allow slightly more generality than $n$ distinct eigenvalues. – Will Jagy May 13 '15 at 20:53
  • @phatty, in the first pair of multiplications, what are the entries in the $M_{12}$ positions for the two different matrix products, the ones with the $pqrs?$ – Will Jagy May 14 '15 at 02:49
  • @WillJagy The first one is $5q$ and the other is $7q$. – 9301293 May 14 '15 at 02:54
  • @phatty, if the two matrices commute, those need to be equal. What does that tell you about $q?$ – Will Jagy May 14 '15 at 02:54
  • @WillJagy $q=0$?... – 9301293 May 14 '15 at 02:55
  • @phatty, yes. Do the $2,1$ position now. – Will Jagy May 14 '15 at 02:56
  • @WillJagy Okay, so I get that it must also be diagonal. But, does that then mean that the $p,q,r,s$ matrix is similar to a diagonal matrix, which Is similar under the same transformation matrix? – 9301293 May 14 '15 at 02:57
  • @phatty, good. In your previous MSE question, replace $A$ by diagonal $D =S^{-1} A S,$ replace $B$ by $F =S^{-1} B S,$ and $C$ by $G =S^{-1} C S$ and fiddle about. – Will Jagy May 14 '15 at 03:05
  • @WillJagy Alright, big moment, I think I got it. If $D=S^{-1}AS, F=S^{-1}BS, G=S^{-1}CS$, then we do some fiddling and end up with the conclusion that $DF=FD$, $DG=GD$. But then both $F, G$ must be diagonal, since the only matrices which commute with diagonal matrices are diagonal, and so, we conclude that $BC=(SFS^{-1})(SGS^{-1})=SFGS^{-1}=SGFS^{-1}=(SGS^{-1})(SFS^{-1})=CB$??? Also, do you know how to link a SE question? I would like to link this one and the previous one I asked in order to help people in the future. – 9301293 May 14 '15 at 03:21
  • @phatty, yes. Note that you need to know that diagonal matrices always commute with each other, which is, well, true. – Will Jagy May 14 '15 at 03:24
  • @WillJagy Yes, I know, I skipped a lot of details. – 9301293 May 14 '15 at 03:25