IMC / 1994 / Problems / Day 2, P10

IMC 1994 · Day 2 · P10

Let $A$ be a $n \times n$ diagonal matrix with characteristic polynomial $(x - c_1)^{d_1} (x - c_2)^{d_2} \dots (x - c_k)^{d_k},$ where $c_1, c_2, \dots, c_k$ are distinct (which means that $c_1$ appears $d_1$ times on the diagonal, $c_2$ appears $d_2$ times on the diagonal, etc. and $d_1 + d_2 + \cdots + d_k = n$ ). Let $V$ be the space of all $n \times n$ matrices $B$ such that $AB = BA$ . Prove that the dimension of $V$ is $d_1^2 + d_2^2 + \cdots + d_k^2.$

Solution (official)

Set $A = (a_{ij})_{i,j=1}^{n}$ , $B = (b_{ij})_{i,j=1}^{n}$ , $AB = (x_{ij})_{i,j=1}^{n}$ and $BA = (y_{ij})_{i,j=1}^{n}$ . Then $x_{ij} = a_{ii} b_{ij}$ and $y_{ij} = a_{jj} b_{ij}$ . Thus $AB = BA$ is equivalent to $(a_{ii} - a_{jj}) b_{ij} = 0$ for $i,j = 1,2,\dots,n$ . Therefore $b_{ij} = 0$ if $a_{ii} \ne a_{jj}$ and $b_{ij}$ may be arbitrary if $a_{ii} = a_{jj}$ . The number of indices $(i,j)$ for which $a_{ii} = a_{jj} = c_m$ for some $m = 1,2,\dots,k$ is $d_m^2$ . This gives the desired result.

IMC 1994 · Day 2 · P10

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