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In mathematics, particularly linear algebra, the Schur product theorem, named after Issai Schur (Schur 1911, p. 14, Theorem VII) (note that Schur signed as J. Schur in Journal für die reine und angewandte Mathematik) states that the Hadamard product of two positive definite matrices is also a positive definite matrix.

Proof

Proof using eigendecomposition

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

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

Each ${\displaystyle (m_{i}\circ n_{j})(m_{i}\circ n_{j})^{T}}$ is positive definite and ${\displaystyle \mu _{i}\nu _{j}>0}$, thus the sum giving ${\displaystyle M\circ N}$ is also psotive definite.

Proof using the trace formula

It is easy to show that for matrices ${\displaystyle M}$ and ${\displaystyle N}$, the Hadamard product ${\displaystyle M\circ N}$ considered as a bilinear form acts on vectors ${\displaystyle a,b}$ as

${\displaystyle a^{T}(M\circ N)b=\operatorname {Tr} (M\operatorname {diag} (a)N\operatorname {diag} (b))}$

where ${\displaystyle \operatorname {Tr} }$ is the matrix trace and ${\displaystyle \operatorname {diag} (a)}$ is the diagonal matrix having as diagonal entries the elements of ${\displaystyle a}$.

Since ${\displaystyle M}$ and ${\displaystyle N}$ are positive definite, we can consider their square-roots ${\displaystyle M^{1/2}}$ and ${\displaystyle N^{1/2}}$ and write

${\displaystyle \operatorname {Tr} (M\operatorname {diag} (a)N\operatorname {diag} (b))=\operatorname {Tr} (M^{1/2}M^{1/2}\operatorname {diag} (a)N^{1/2}N^{1/2}\operatorname {diag} (b))=\operatorname {Tr} (M^{1/2}\operatorname {diag} (a)N^{1/2}N^{1/2}\operatorname {diag} (b)M^{1/2})}$

Then, for ${\displaystyle a=b}$, this is written as ${\displaystyle \operatorname {Tr} (A^{T}A)}$ for ${\displaystyle A=N^{1/2}\operatorname {diag} (a)M^{1/2}}$ and thus is positive. This shows that ${\displaystyle (M\circ N)}$ is a positive definite matrix.

Proof using Gaussian integration

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

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

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

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

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 ${\displaystyle \langle X_{i}X_{j}\rangle =M_{ij}}$, ${\displaystyle \langle Y_{i}Y_{j}\rangle =N_{ij}}$ and independt from each other so that we have

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

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

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

Using Wick's theorem to develop

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

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

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

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

Refercenes

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