How to find eigenvalues of following symmetric matrix
$\begin{bmatrix} kI-A & -A & -A & \cdots & -A\\ -A & kI-A & -A & \cdots & -A\\ -A & -A & kI-A & \cdots & -A\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ -A & -A & -A & \cdots & kI-A\\ \end{bmatrix}_{m+1}$
Where, $A$ is any square matrix of order $n$,
$I$ is identity matrix of order $n$,
$k \in \mathbb{N}$
As well as one eigenvalue of above matrix is zero.
I think if we can convert above matrix in terms of either Kronecker product or Kronecker sum of two matrices then we can find eigenvalues of above matrix by taking multiplication or addition of two matrices respectively.
The other way is might be if we can convert above matrix in block diagonal matrix then we can find eigenvalues easily.
Even if small hint will also help me to solve the problem
First remark, your matrix is not symmetric unless $A$ is so. For simplicity I will assume at least that $A$ is diagonalisable.
Let $B$ be the matrix obtained for $k=0$; it is the Kronecker product of an $m'\times m'$ matrix$~P$ with all its entries equal to$~1$, and $-A$, for $m'=m+1$ (what was the point of the shift in $m$ anyway?). $P$ has rank$~1$ and trace$~m'$, so it has eigenvalue $0$ with multiplicity (geometric and algebraic) $m'-1$, one a simple eigenvalue$~m'$. Then $B=P\otimes-A$ has eigenvalue $0$ with multiplicity $(m'-1)n$, together with the multiset of eigenvalues of$~A$ each multiplied by$~{-}m'$ (if $0$ happens to be an eigenvalue of$~A$, then the multiplicity of $0$ will actually exceed $(m'-1)n$).
Now the given matrix is $B+kI$, and has the same eigenvalues shifted up by$~k$: eigenvalue$~k$ with multiplicity $(m'-1)n$, together with eigenvalues $k-m'\lambda$ for each eigenvalue$~\lambda$ of$~A$.