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

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

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?

Leox
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  • mostly you calculate a matrix with columns the eigenvectors, adjusting so they are all integers, and see if there is any pattern that holds up as $n$ increases – Will Jagy Dec 04 '18 at 22:40
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    Your matrix, say $M$, belongs to the so-called category of "skew centrosymmetric matrices" $JMJ=-M$ (where $J$ is the antidiagonal matrix). Doesn't know if that helps... – Jean Marie Dec 04 '18 at 23:50

2 Answers2

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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 $d_{kk}=(-1)^{\lfloor(k-1)/2\rfloor}$ (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 $\pm\lambda$. 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 $2$.

user1551
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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 :

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

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.

Jean Marie
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