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If A is a fixed n by n matrix, and S := {$B\in M_n(\Bbb{C}) : AB = BA$} is a subspace of $M_n(\Bbb{C})$.

Find a 3 by 3 matrix A such that S has dimension 5.

I figured the matrix below has dimension 3 when multiplied with B

$$A= \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & i \\ \end{pmatrix}$$

So for a general matrix B, I did AB=BA, and then compared both sides and wherever the corresponding entries didn't match, I concluded them to equal zero. $$ B= \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & j \\ \end{pmatrix}$$

user117498
  • 775
  • you need to choose a,b,c,d,... so that the corresponding entries do match. you will find out that everything is zero except a,e, and j. so $S$ has dimension $3$ not $5$ as required – Lozenges Mar 20 '15 at 19:25
  • I need to find a matrix A such that when multiplied by B on either side it has dimension 5. The matrix A above is what I tried that gave me dimension 3, not 5. Thus, it's wrong. So the goal is to find a fixed matrix A that has dimension 5. – user117498 Mar 21 '15 at 11:04
  • that is right. I gave an example of such a matrix in my answer below. First, verify that $AB=BA$ – Lozenges Mar 22 '15 at 11:58

1 Answers1

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Try the matrix

A= \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} and

B= \begin{pmatrix} a & b & c \\ 0 & a & 0 \\ 0 & h & j \\ \end{pmatrix}

Lozenges
  • 1,742
  • May I ask how you came up to this solution. – user117498 Mar 23 '15 at 02:13
  • when the minimal polynomial coincides with the characteristic polynomial then a matrix $A$ commutes only with $p(A)$ where $p$ is a polynomial. see http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it so I looked for a matrix whose minimal polynomial is $(x-1)^2$ – Lozenges Mar 23 '15 at 15:38