Largest and smallest eigenvalues of matrix with alternating 1s and 0s

Question

My linear algebra is a bit rusty, so this may be a dumb question. But suppose I have an $N$-by-$N$ block matrix $A$, where each $(i,j)$ entry $A_{ij}$ is the $2\times 2$ identity matrix. Equivalently, $A={(a_{mn})}$, where $A$ is the $2N\times2N$ matrix with entries $a_{mn}=1$ if $m+n$ is even and $a_{mn}=0$ if $m+n$ is odd.

I am interested in the largest $\lambda_{max}$ and smallest $\lambda_{\min}$ eigenvalues of $A$. From the Gershgorin circle theorem, we have that $\lambda_{\max}\leq N$. Also, if $e_{i}$ denotes the $i$-th standard basis vector, then $\sum_{i: \text{$i$ odd}} e_i$ is an eigenvector with eigenvalue $N$, so $\lambda_{max}=N$. I suspect that $A$ is positive semi-definite, so that $\lambda_{min}=0$; but I don't know how to prove this.

Answer 1

One can work out the eigenvalues exactly in fact:

• The eigenvalue $N$ occurs with multiplicity $2$, with corresponding eigenvectors $$ (1,0,1,0,\dots,1,0) \quad\text{and}\quad(0,1,0,1,\dots,0,1). $$
• The eigenvalue $0$ occurs with multiplicity $2N-2$, with corresponding eigenvectors $$ (1,0,-1,0,\dots,0),\, (0,1,0,-1,0,\dots,0),\, (0,0,1,0,-1,0,\dots,0),\, \dots, (0,\dots,0,1,0,-1). $$ Alternately, the fact that there are only two distinct rows means that the matrix has rank at most $2$ (and those two rows being linearly independent means the rank is exactly $2$), which also implies that the eigenvalue $0$ has multiplicity $2N-2$ by the rank-nullity theorem.