Why is this not a valid Variance Covariance matrix, and inherently not positive semi-definite?

Question

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]  1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[2,] -0.5  1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[3,] -0.5 -0.5  1.0 -0.5 -0.5 -0.5 -0.5 -0.5
[4,] -0.5 -0.5 -0.5  1.0 -0.5 -0.5 -0.5 -0.5
[5,] -0.5 -0.5 -0.5 -0.5  1.0 -0.5 -0.5 -0.5
[6,] -0.5 -0.5 -0.5 -0.5 -0.5  1.0 -0.5 -0.5
[7,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5  1.0 -0.5
[8,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5  1.0

The reason I know is because this results in a negative eigenvalue, and variance-covariance matrices are positive semi-definite.

My thinking here was to have the variances be one, so that the correlations were the covariances, and thus equal -0.5.

Is there something with the theory I am missing? I understand that it is not positive semi-definite and how to show as such, but I am more curious what assumptions this is violating in terms of probability/statistics.

I went to generate MVN data with this variance-covariance structure and realized this wasn't positive semi-definite, and then became curious what was inherently wrong with this matrix.

Answer 1

As shown here, an $n\times n$ matrix with $a$ on the diagonal and $b$ elsewhere, that is, the matrix $$\begin{bmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{bmatrix}$$ has $\color{blue}{a +(n-1)b}$ as one of its eigenvalues. For your particular matrix, we have $a = 1, b = -0.5, n =8$, so $$a + (n-1)b = 1 + 7\times (-0.5) = -2.5< 0.$$ Hence your symmetric matrix has a negative eigenvalue ($-2.5$), so cannot be positive semi-definite. This implies that it is not a valid variance-covariance matrix.

EDIT: Just realised you said you already know this...