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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 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

Let ${\displaystyle X}$ be an ${\displaystyle n}$ dimensional centered Gaussian random variable with covariance ${\displaystyle \langle X_{i}X_{j}\rangle =C_{ij}}$. 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

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

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

Refercenes

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