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 $V$ ?


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.

  • $\begingroup$ 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? $\endgroup$ – user1551 Nov 6 '18 at 11:32
  • $\begingroup$ But would partitioning not make it a little messy ? $\endgroup$ – Yadati Kiran Nov 6 '18 at 11:38
  • $\begingroup$ Could you please elaborate? $\endgroup$ – Yadati Kiran Nov 6 '18 at 11:40
  • $\begingroup$ 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 $B_{ij}$. $\endgroup$ – user1551 Nov 6 '18 at 11:45

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