A question from entrance test of PRIMES 2016, namely M3.
The solution says:
The matrix has eigenvalues $0, 1, ...., p-1$ with eigenspaces of dimension $n_0, n_1, ...n_{p-1}$. The group $GL_n(\mathbb F_p)$ acts on such arrangements transitively, with the stabilizer $GL_{n_0}(\mathbb F_p) \times GL_{n_1}(\mathbb F_p) \times ..... \times Gl_{n_{p-1}}(\mathbb F_p)$. So the number of matrices is : $$\sum_{{n_0},...,{n_{p-1}}:\sum n_i =n} \frac{\prod_{j=0}^{n-1} (p^n - p^j)}{\prod_{i=0}^{p-1} \prod_{j=0}^{n_i -1} (p^{n_i -1} -p^j)}$$
So, could anybody help me understand why is the product of bunch of general linear group $GL_{n_0}(\mathbb F_p) \times GL_{n_1}(\mathbb F_p) \times ..... \times Gl_{n_{p-1}}(\mathbb F_p)$ the stabilizer? And I can't really see the importance of eigenspace here in this proof..
Besides, what is it summing there... it's $\sum n_i =n$ in the right corner under that bigger sigma in case that equal sign is not clear.
Part of the argument is that $A$ is diagonalizable (with eigenvalues in $\mathbb F_p$). One way to see this is using the following identity in the ring of polynomials $\mathbb F_p[x]$: $$ \prod_{a\in\mathbb F_p}(x-a)=x^p-x. $$ It follows that $$ \prod_{a\in\mathbb F_p}(A-aI_n)=0. $$ This implies that $$ \mathbb F_p^n=\bigoplus_{a\in\mathbb F_p}V_a $$ where $V_a=\ker(A-aI_n)$ is the $a$-eigenspace of $A$. Since $A$ acts as $a$ on $V_a$, $A$ is determined by the spaces $V_0,\ldots,V_{p-1}$. So the formula is counting the number of such sequences of subspaces. The sum is over the possible dimensions of the subspaces.