Ytukyg

Find eigenvalues of the $(n+1) times (n+1)$-matrix

$$ left( begin {array}{cccccccc} 0&0&0&0&0&0&-n&0\ 0
&0&0&0&0&-(n-1)&0&1\ 0&0&0&0&-(n-2)&0&2&0
\ 0&0&0&ldots&0&3&0&0\ 0&0&-3&0&ldots&0
&0&0\ 0&-2&0&n-2&0&0&0&0\ -1&0&n-1&0
&0&0&0&0\ 0&n&0&0&0&0&0&0end {array} right)
$$

I have tried for some small $n$ by hand and have got that for $n=3$ the eigenvalues are $1,-1,3,-3$, for $n=4$ eigenvalues are $0,2,-2,4,-4$. So I am conjectured than for arbitrary $n$ the eigenvalues are $n, n-2, n-4, ldots, -n+2,-n.$

How to prove it?

Call your matrix $B_{n+1}$. Denote the Kac matrix of the same size by
$$
A_{n+1}=pmatrix{
0&n\
1&0&n-1\
&2&ddots&ddots\
& &ddots&ddots&ddots\
& & &ddots&0 &1\
& & & & n &0}.
$$
It is known that the spectrum of the Kac matrix is given by
$$
sigma(A_{n+1})={-n,,-n+2,,-n+4,ldots,,n-4,,n-2,,n}.
$$
We shall prove that $sigma(A_{n+1})=sigma(B_{n+1})$. Consequently (as the Kac matrix has distinct eigenvalues) the two matrices are similar to each other.

Here is our proof. First, $B_{n+1}^2$ is similar to $A_{n+1}^2$. In fact, if $D$ denotes the $(n+1)times(n+1)$ diagonal matrix such that $d_{kk}=(-1)^{lfloor(k-1)/2rfloor}$ (i.e. $D=operatorname{diag}(1,1,-1,-1,1,1,-1,-1,ldots)$), a straightforward computation will show that $DB_{n+1}^2D=A_{n+1}^2$.

Thus $sigma(B_{n+1}^2)=sigma(A_{n+1}^2)$. Yet, as pointed out in the answer by Jean Marie, $B_{n+1}$ is similar to $-B_{n+1}$ via the reversal matrix (the mirror image of the identity matrix from left to right). Therefore, nonzero eigenvalues of $B_{n+1}$ must occur in pairs of the form $pmlambda$. Hence we have $sigma(B_{n+1})=sigma(A_{n+1})$, because the multiplicity of each nonzero eigenvalue of $A_{n+1}^2$ is exactly $2$.

This is very far from a complete answer (but too long to fit in a comment).

Let us call $M$ (or $M_n$ if we want to stress the dependency on $n$) the given matrix.

Let us establish the following property : if $lambda$ is an eigenvalue of $M$, then $-lambda$ is as well an eigenvalue of $M$. Said otherwise, the spectrum of $M$ is symmetric with respect to $O$.

Let $J$ be the antidiagonal matrix

($J_{ij}=1 iff i+j=n+2$ and $J_{ij}=0$ otherwise).

Please note that $J^{-1}=J$.

It is not difficult to establish that

$$JMJ=-M. tag{1}$$

This is a way to express that matrices like $M$ are "skew-centrosymmetric" (there is some literature on the subject).

Let $I$ be the $n+1$ dimensional unit matrix. Subtracting $lambda I$ to LHS and RHS of (1), one can write :

$$JMJ-lambda JIJ=-M-lambda I $$

Let us left- and -right factorize by $J$ :

$$J(M - lambda I)J=-(M+lambda I).$$

Taking determinants of both sides, we get :

$$det(J)^2det(M - lambda I)=(-1)^{n+1}det(M+lambda I).$$

Thus, as $det(J)neq 0$ (recall that $J^2=I$) :

$$det(M - lambda I)=0 iff det(M-(-lambda) I)=0,$$

proving the result.