# Dimension of vector space of matrices simltaneously diagonalizable with A.

Let $$A$$ be a $$55 \times 55$$ diagonal matrix with characteristic polynomial $$(x - c_1)(x - c_2)^2(x - c_3)^3\cdots(x - c_{10})^{10}$$ where $$c_1,\cdots,c_{10}$$ are all distinct. Let $$V$$ be the vector space of all $$55 \times 55$$ matrices $$B$$ such that $$AB = BA$$. What is the dimension of $$V$$ ?

Try:

Since $$A$$ is given to be diagonal matrix and $$AB = BA \implies A$$ and $$B$$ share same eigenbasis, hence are simultaneously diagonalizable. $$\exists\, P$$ non singular such that $$P^{-1}AP$$ and $$P^{-1}BP$$ are diagonal. Using the map $$f:V \rightarrow A$$ as $$f(B)=P^{-1}BP \quad \forall B\in V$$, we get dimension of $$V$$ as 10.

• In the worst case, try a coordinate-ful proof. Let $A=\operatorname{diag}(c_1,c_2,c_2,c_3,c_3,c_3,\ldots,c_{10})$ and partition $B$ into a block matrix. If $AB=BA$, what do the sub-blocks of $B$ look like? – user1551 Nov 6 '18 at 11:32
• But would partitioning not make it a little messy ? – Yadati Kiran Nov 6 '18 at 11:38
• Could you please elaborate? – Yadati Kiran Nov 6 '18 at 11:40
• Let $A=c_1I_1\oplus c_2I_2\oplus\cdots\oplus c_{10}I_{10}$. Partition $B$ as a $10\times10$ block matrix with conforming partition to $A$, so that the $k$-th diagonal sub-block of $B$ is $k\times k$ and the size of the $(i,j)$-th sub-block, denoted by $B_{ij}$, is $i\times j$. The equation $AB=BA$ then puts some constraints on each $B_{ij}$. – user1551 Nov 6 '18 at 11:45