Is a symmetric matrix with a positive determinant a covariance matrix?

Question

A covariance matrix has a positive determinant (the determinant is equal to the product of the eigenvalues which are all positive).

Is the other direction also true? That is, if a symmetric real-valued matrix has a a positive determinant, is it then a covariance matrix?

Answer 1

From the spectral theorem any positive-semidefinate matrix $C$ has a square root $S:=C^{1/2}$ matrix such that $C=SS^T$. Now let $X=(X_1,\cdots,X_n)$, where $P(X_i=\pm 1)=1/2$, so that:

$$E[(SX-E[SX])(SX-E[SX])^T]=E[SXX^TS^T]=SS^T=C.$$

Note that positive-semidefinite implies positive determinant, but is also stronger as it guarantees all eigenvalues are non-negative. blueInk's example in the comments above has negative eigenvalues.