For two square matrices $A$ and $B$ of the same size $n$ let us define $A★B$ following matrix:
$ (A★B)_{ij} = \begin{cases} (AB)_{ij}, & \text{if $i$ is odd,} \\ b_{ij}, & \text{else.} \end{cases}$
For matrix $A$ let us define operator $\Phi_A : B \mapsto A \bigstar B$ on the $ n \times n $ matrices.
(a) Can matrix $A$ exist so this operator eigenvalue is $2$?
(b) What is the maximum number of eigenvalues this operator can have (for fixed n)?
I solved (a):
Yes, for $ A = 2 I $, where $I$ is identity matrix, and $ (B)_{ij} = \begin{cases} 1, \text{if i is odd}, \\ 0, \text{else}. \end{cases}$
I am unsure how to solve b and would appreciate a hint. Thanks very much!
Hint. Let $(r,s)=(n,n-1)$ when $n$ is odd, or $(n-1,n)$ when $n$ is even. That is, $r$ and $s$ are respectively the last odd index and the last even index in $\{1,2,\ldots,n\}$.
Let $E_{ij}\in M_{n,n}$ be the matrix with a $1$ at the $(i,j)$-th position and zeroes elsewhere. Consider the ordered basis $\mathcal B$ of $M_{n,n}$ consisting of the $E_{ij}$s in the following order: \begin{aligned} &E_{11},E_{31},\ldots,E_{r1},E_{21},E_{41},\ldots,E_{s1},\\ &E_{12},E_{32},\ldots,E_{r2},E_{22},E_{42},\ldots,E_{s2},\\ &\ldots,\\ &E_{1n},E_{3n},\ldots,E_{rn},E_{2n},E_{4n},\ldots,E_{sn}.\\ \end{aligned} The matrix representation of the linear operator $L(B)=A★B$ with respect to $\mathcal B$ is then $$ M=\pmatrix{X&Y\\ 0&I_{s/2}\\ &&X&Y\\ &&0&I_{s/2}\\ &&&&\ddots\\ &&&&&X&Y\\ &&&&&0&I_{s/2}} $$ where $\pmatrix{X&Y}$ is the submatrix consisting of the odd-indexed rows of $A$.