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Suppose I have a $n \times n$ symmetric matrix along the main diagonal that is also anti-symmetric along the other diagonal. The rows and columns add to 0. For example,

$$ \begin{bmatrix} a & b & c & 0 \\ b & d & 0 & -c \\ c & 0 & -d & -b \\ 0 & -c & -b & -a \end{bmatrix} $$

I want to show for the general case that all eigenvalues are zero except two, one positive and one negative, equal in magnitude but opposite. My idea is to show the rank of the matrix is only $2$, which then since the trace is zero I can say that the two non-zero eigenvalues must be opposites.

The problem however is that I can't show rank $=2$ for a general $n \times n$. All I can reason is that since it is a Zero Line Sum matrix, the rank is at most $n-1$. Perhaps I am missing a condition that could make this true? Any help is appreciated. Thanks!

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  • $\begingroup$ Let $T$ be the matrix with $T_{ij}=1$ if $i+j=n$ and $T_{ij}=0$ otherwise. Let $M$ be your matrix. Maybe one can use $TM+MT=0$? $\endgroup$
    – tkf
    Commented Jul 3, 2020 at 2:15
  • $\begingroup$ You're right, that holds here. I did some digging and the matrix I'm talking about is called a skew-centrosymmetric matrix (defined by what you stated). As far as using the row/colum sum property to show the rank=2, im still stuck there unfortunately. $\endgroup$
    – Alex B
    Commented Jul 3, 2020 at 2:43
  • $\begingroup$ I think it might be clearer in the basis $(100...001), (010...010), (001...100),..., (100...00-1), (010...0-10), (001...-100)$. $\endgroup$
    – tkf
    Commented Jul 3, 2020 at 3:28

1 Answer 1

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The statement you are trying to prove isn't true when $n \ge 5$. For example, when $n = 5$, consider the matrix $$\begin{bmatrix}1&0&0&-1&0 \\ 0&0&-1&0&1 \\ 0&-1&0&1&0 \\ -1&0&1&0&0 \\ 0&1&0&0&-1\end{bmatrix}.$$

You can check that the above matrix satisfies all the desired properties, but its rank is $4$ not $2$.

In general, if we define an $n \times n$ matrix $A$ by $$A_{i,j} = \begin{cases}1 & \text{if} \ i=j=1 \\ -1 & \text{if} \ i+j=n \\ 1 & \text{if} \ i+j = n+2 \\ -1 & \text{if} \ i=j=n \\ 0 & \text{otherwise}\end{cases},$$ then $A$ satisfies the conditions of the problem, but $\text{rank}(A) = \begin{cases}n-1 & \text{if} \ n \ \text{is odd} \\ n-2 & \text{if} \ n \ \text{is even}\end{cases}.$

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