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Let $D$ be an $n \times n$ diagonal matrix whose distinct diagonal entries are $d_1,\ldots, d_k$, and where $d_i$ occurs exactly $n_i$ times. For the subspace $W$ of $M_{n \times n}(F)$ defined by $W=\{A : AD = DA\}$ prove that $\text{dim}(W) = n_1^2 + n_2^2 +\ldots+ n_k^2$

I know that $A \in W$ must be symmetric, and I can see that if each $d_k$ is distinct and only occurs once, that $W$ has dimension $n$. I also realize that if $D_{ii} = D_{jj}$ then $A_{ji}$ can be anything and $A_{ij}$ can be anything. (that's terrible wording but this question has me so lost), and if $D_{ii} \neq D_{jj}$ then $A_{ij} = 0$ and $A_{ji} = 0$.

I'm not sure how to formalize any of my thoughts on this question at all. It is a homework question that I'd like to understand fully.

Trigginometric
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1 Answers1

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let $$ D = \left( \begin{array}{cccc} 5&0&0&0 \\ 0&5&0&0 \\ 0&0&7&0 \\ 0&0&0&7 \end{array} \right) $$ and $$ A = \left( \begin{array}{cccc} a&b&c&d \\ e&f&g&h \\ i&j&k&l \\ m&n&o&p \end{array} \right) $$ Write out $AD$ and $DA$ and say EXACTLY what matrices $A$ commute with $D$

Will Jagy
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  • I've done several different cases, and see that the pattern fits quite easily, but just have no idea how to actually prove the statement – Trigginometric Oct 26 '18 at 00:24
  • What am I supposed to be taking away from the individual cases? – Trigginometric Oct 26 '18 at 00:29
  • @Trigginometric what are the conditions on the entries of $A$ (the 16 letters) that make $AD=DA?$ What letters need to be zero, and what are the remaining letters that are allowed to be anything? What pattern do those make, the nonzero ones? – Will Jagy Oct 26 '18 at 00:33
  • c=d=g=h=i=j=m=n=0 and the others can be what they want – Trigginometric Oct 26 '18 at 00:35
  • @Trigginometric good. Now draw a copy of $A$ with all those set to zero that need to be zero. There are block shapes – Will Jagy Oct 26 '18 at 00:36
  • But what if $D$ doesn't have identical values directly diagonal from one another? I tried doing it that way and got non-block shapes but the statement still holds. I'm just not sure how to turn that information into a proof – Trigginometric Oct 26 '18 at 00:40
  • One approach is as follows: if the diagonal entries aren't grouped together, then there is a permutation matrix $P$ such that $PDP^T$ does have its diagonal entries grouped together. – Ben Grossmann Oct 26 '18 at 00:43
  • Another way to go is to find a formula for the $i,j$ entries of $AD$ and $DA$, which is easier than it sounds – Ben Grossmann Oct 26 '18 at 00:45
  • @Trigginometric please do the same thing, this time with a 7 by 7 matrix $D$ in the upper left corner a 3 by 3 block with diagonal all 5, then the lower right corner a 4 by 4 block with all diagonal 7. Please do not jump to permuted matrices yet. – Will Jagy Oct 26 '18 at 00:48