Why is this sum of Kronecker products singular?

Question

Consider the $2n\times 2n$ matrix $$M(n):=I_2\otimes I_n+(J_2-I_2)\otimes(J_n-I_n),$$ where $J_n$ denotes the $n\times n$ matrix whose entries are all $1$s, $I_n$ is the identity, and $\otimes$ is the Kronecker product.

For example, when $n=3$ we have $$M(3)=\left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ \hline 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ \end{array} \right)\!.$$

I have found that $\det M(n)$ appears to be $0$ for all positive integer values of $n\geq2$. I have tested this conjecture for $n\leq 200$ without counterexamples.

Q. Why is $\det M(n)=0$ for $n\geq2$?

If we expand the second term in the definition of $M(n)$, we get $$M(n)=2(I_2\otimes I_n)+J_2\otimes J_n-J_2\otimes I_n-I_2\otimes J_n.$$ From here, I don't see a way to simplify the expression for $M(n)$. Of course, no formula for the determinant of the sum of two matrices exists, so we cannot hope to simplify $\det M(n)$ directly along these lines. We can also observe that $I_2\otimes I_n=I_{2n}$, but really that does not help at all.

What might be a way forward? Thanks in advance!

Answer 1

Add Row1 to Row4 and Row2 to Row5.

These are elementary row operations that don't change the determinant, but now Row4 and Row5 are equal, so the determinant is zero.

Answer 2

$M(n)$ is a real symmetric matrix (hence diagonalizable) that is similar to $diag(0_{n-1},2I_{n-1},-n+2,n)$

EDIT. Answer to the OP. Indeed, $spectrum(J_k-I_k)=\{(k-1)\times -1,k-1\}$; on the other hand, if $spectrum(U)=(\lambda_i)_{i\leq p},spectrum(V)=(\mu_j)_{j\leq q}$, then, $spectrum(U\otimes V)=(\lambda_i\mu_j)_{i\leq p,j\leq q}$.

Answer 3

Let $n\ge2$, $e=(1,1)^T$ and $0\ne v\in\ker(J_n)$. Then $M(n)$ is singular, because $$ M(n)(e\otimes v) =e\otimes v+\left[(J_2-I_2)e\right]\otimes\left[(J_n-I_n)v\right] =e\otimes v+e\otimes-v=0. $$