Case of M = N

Let ${\displaystyle X}$ be an ${\displaystyle n}$-dimensional centered Gaussian random variable with covariance ${\displaystyle \langle X_{i}X_{j}\rangle =M_{ij}}$. Then the covariance matrix of ${\displaystyle X_{i}^{2}}$ and ${\displaystyle X_{j}^{2}}$ is

${\displaystyle \operatorname {Cov} \left(X_{i}^{2},X_{j}^{2}\right)=\left\langle X_{i}^{2}X_{j}^{2}\right\rangle -\left\langle X_{i}^{2}\right\rangle \left\langle X_{j}^{2}\right\rangle }$

Using Wick's theorem to develop ${\displaystyle \left\langle X_{i}^{2}X_{j}^{2}\right\rangle =2\left\langle X_{i}X_{j}\right\rangle ^{2}+\left\langle X_{i}^{2}\right\rangle \left\langle X_{j}^{2}\right\rangle }$ we have

${\displaystyle \operatorname {Cov} \left(X_{i}^{2},X_{j}^{2}\right)=2\left\langle X_{i}X_{j}\right\rangle ^{2}=2M_{ij}^{2}}$

Since a covariance matrix is positive definite, this proves that the matrix with elements ${\displaystyle M_{ij}^{2}}$ is a positive definite matrix.

General case

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be ${\displaystyle n}$-dimensional centered Gaussian random variables with covariances ${\displaystyle \left\langle X_{i}X_{j}\right\rangle =M_{ij}}$, ${\displaystyle \left\langle Y_{i}Y_{j}\right\rangle =N_{ij}}$ and independent from each other so that we have

${\displaystyle \left\langle X_{i}Y_{j}\right\rangle =0}$ for any ${\displaystyle i,j}$

Then the covariance matrix of ${\displaystyle X_{i}Y_{i}}$ and ${\displaystyle X_{j}Y_{j}}$ is

${\displaystyle \operatorname {Cov} \left(X_{i}Y_{i},X_{j}Y_{j}\right)=\left\langle X_{i}Y_{i}X_{j}Y_{j}\right\rangle -\left\langle X_{i}Y_{i}\right\rangle \left\langle X_{j}Y_{j}\right\rangle }$

Using Wick's theorem to develop

${\displaystyle \left\langle X_{i}Y_{i}X_{j}Y_{j}\right\rangle =\left\langle X_{i}X_{j}\right\rangle \left\langle Y_{i}Y_{j}\right\rangle +\left\langle X_{i}Y_{i}\right\rangle \left\langle X_{j}Y_{j}\right\rangle +\left\langle X_{i}Y_{j}\right\rangle \left\langle X_{j}Y_{i}\right\rangle }$

and also using the independence of ${\displaystyle X}$ and ${\displaystyle Y}$, we have

${\displaystyle \operatorname {Cov} \left(X_{i}Y_{i},X_{j}Y_{j}\right)=\left\langle X_{i}X_{j}\right\rangle \left\langle Y_{i}Y_{j}\right\rangle =M_{ij}N_{ij}}$

Since a covariance matrix is positive definite, this proves that the matrix with elements ${\displaystyle M_{ij}N_{ij}}$ is a positive definite matrix.

Proof of positive semidefiniteness

Let ${\displaystyle M=\sum \mu _{i}m_{i}m_{i}^{\textsf {T}}}$ and ${\displaystyle N=\sum \nu _{i}n_{i}n_{i}^{\textsf {T}}}$. Then

${\displaystyle M\circ N=\sum _{ij}\mu _{i}\nu _{j}\left(m_{i}m_{i}^{\textsf {T}}\right)\circ \left(n_{j}n_{j}^{\textsf {T}}\right)=\sum _{ij}\mu _{i}\nu _{j}\left(m_{i}\circ n_{j}\right)\left(m_{i}\circ n_{j}\right)^{\textsf {T}}}$

Each ${\displaystyle \left(m_{i}\circ n_{j}\right)\left(m_{i}\circ n_{j}\right)^{\textsf {T}}}$ is positive semidefinite (but, except in the 1-dimensional case, not positive definite, since they are rank 1 matrices). Also, ${\displaystyle \mu _{i}\nu _{j}>0}$ thus the sum ${\displaystyle M\circ N}$ is also positive semidefinite.

Proof of definiteness

To show that the result is positive definite requires even further proof. We shall show that for any vector ${\displaystyle a\neq 0}$, we have ${\displaystyle a^{\textsf {T}}(M\circ N)a>0}$. Continuing as above, each ${\displaystyle a^{\textsf {T}}\left(m_{i}\circ n_{j}\right)\left(m_{i}\circ n_{j}\right)^{\textsf {T}}a\geq 0}$, so it remains to show that there exist ${\displaystyle i}$ and ${\displaystyle j}$ for which corresponding term above is nonzero. For this we observe that

${\displaystyle a^{\textsf {T}}(m_{i}\circ n_{j})(m_{i}\circ n_{j})^{\textsf {T}}a=\left(\sum _{k}m_{i,k}n_{j,k}a_{k}\right)^{2}}$

Since ${\displaystyle N}$ is positive definite, there is a ${\displaystyle j}$ for which ${\displaystyle n_{j}\circ a\neq 0}$ (since otherwise ${\displaystyle n_{j}^{\textsf {T}}a=\sum _{k}(n_{j}\circ a)_{k}=0}$ for all ${\displaystyle j}$), and likewise since ${\displaystyle M}$ is positive definite there exists an ${\displaystyle i}$ for which ${\displaystyle \sum _{k}m_{i,k}(n_{j}\circ a)_{k}=m_{i}^{\textsf {T}}(n_{j}\circ a)\neq 0.}$ However, this last sum is just ${\displaystyle \sum _{k}m_{i,k}n_{j,k}a_{k}}$. Thus its square is positive. This completes the proof.