Find the dimension of the vector space V.

Question

Let $A$ be a $11\times 11$ diagonal matrix with the characteristic polynomial $(x-3)^4(x-4)(x-7)^4(x-9)^2$. Let $V$ be the space of $11\times 11$ matrices such that $AB=BA$. Then find the dimension of $V$.

For doing such questions I always considered one general matrix with variable entries and used to get certain conditions on them. Here things won't seems easy that way.I know that dimension of $V$ is nothing but the null space of $(AB-BA)$ also I am aware of the relation $Trace(AB)=Trace(BA)$, or something is hidden in the question itself that I can't see! Couldn't think further, the correct answer given is 37.

Thanks.

Answer 1

Let us observe that a diagonal $A$ with diagonal entries $d_1,d_2,\dots, d_{11}$ has characteristic polynomial $P_A(\lambda)=(\lambda - d_1)(\lambda-d_2)\dots(\lambda-d_{11})$, so in our case we have on the diagonal the elements $3,3,3,3;4;7,7,7,7;9,9$ - possibly in some other order. After a conjugation with a permuation matrix we may and do assume that the diagonal entries are exactly in this order, so $A$ is written as a block matrix $11\times 11$ splitted in blocks $$ (4+1+4+2)\times(4+1+4+2) $$ as follows: $$ A=\begin{bmatrix} 3 \\ & 4 \\ && 7\\&&&9 \end{bmatrix} $$ Now consider a matrix $B$ in the same block format, $$ B=\begin{bmatrix} B_{11} & B_{12} & B_{13} & B_{14} \\ B_{21} & B_{22} & B_{23} & B_{24} \\ B_{31} & B_{32} & B_{33} & B_{34} \\ B_{41} & B_{42} & B_{43} & B_{44} \\ \end{bmatrix} $$ and compute $AB$ and $BA$, and compare. (We have multiplication with $3,4,7,9$ on lines, respectively on columns.) It turns out that $AB=BA$ is equivalent to the vanishing of all off-diagonal blocks, so $$ B=\begin{bmatrix} B_{11} \\ & B_{22} \\ & & B_{33} \\ & & & B_{44} \\ \end{bmatrix} $$ is a matrix with $4^2+1^2+4^2+2^2=16+1+16+4=37$ degrees of freedom.

Answer 2

Suppose $A$ is a diagonal matrix with diagonal entries $a_i$ and let $B =(b_{ij})$. Then $AB = BA$ $\iff$ $a_ib_{ij} = a_jb_{ij}$ for all $i$, $j$. Thus when $a_i \not= a_j$, $b_{ij}$ must be zero. Whereas if $a_i = a_j$ then $b_{ij}$ can take any value. That gives you the dimension you need. The number of pairs $(i, j)$ such that $a_i = a_j$. If some value is repeated $n$ times in the diagonal then you have $n^2$ pairs satisfying that. Now you should get why 37 is the answer.