# Find eigenvalues of the matrix

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?

• 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 4 '18 at 22:40
• 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 4 '18 at 23:50

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 $$2$$.

• Very clever manipulations... – Jean Marie Dec 6 '18 at 20:52
• @user1551 Thank you!! – Leox Dec 7 '18 at 18:46

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.