Is there any intuition why the following matrix is positive semidefinite

linear algebramatricespositive-semidefinitesymmetric matrices

I have the following 8 by 8 square matrix, which is positive semidefinite:
\begin{bmatrix}3&1&1&-1&1&-1&-1&-3\\1&3&-1&1&-1&1&-3&-1&
\\ 1&-1&3&1&-1&-3&1&-1 \\ -1&1&1&3&-3&-1&-1&1
\\ 1&-1&-1&-3&3&1&1&-1 \\ -1&1&-3&-1&1&3&-1&1
\\ -1&-3&1&-1&1&-1&3&1 \\ -3&-1&-1&1&-1&1&1&3
\end{bmatrix}

I wonder if there is an intuitive argument why this matrix is positive semidefinite.

This matrix has some interesting properties:

  1. It is symmetric, diagonal elements have the largest values, and antidiagonal have the smallest values.

  2. The sum of each row and column is $0$.

  3. I also notice it is symmetric in the sense that first column is the inverse of the last column, the second column is the inverse of the 7th column.

If it helps, I computed the eigenvalues, which are $0,0,0,0,0,8,8$ and $8$.

Best Answer

If $A$ is the $4\times 4$ submatrix in the upper left corner and $J$ is the negative permutation matrix $$ J=\begin{bmatrix} &&&-1\\ &&-1&\\ &-1&&\\ -1&&& \end{bmatrix} $$ then the original matrix is $$ \begin{bmatrix} A & AJ\\ JA & JAJ \end{bmatrix}= \begin{bmatrix} I\\J\end{bmatrix}A \begin{bmatrix} I &J\end{bmatrix} $$ which is positive semidefinite if $A$ is. The positive semidefiniteness of $A$ follows e.g. from $$ A=2(I+J)+ee^T $$ where $e$ is the vector of all ones.

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