Schur product theorem

Template:Short description In mathematics, particularly in linear algebra, the Schur product theorem states that the Hadamard product of two positive definite matrices is also a positive definite matrix. The result is 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.)

The converse of the theorem holds in the following sense: if $M$ is a symmetric matrix and the Hadamard product $M\circ N$ is positive definite for all positive definite matrices $N$, then $M$ itself is positive definite.

Proof

Proof using the trace formula

For any matrices $M$ and $N$, the Hadamard product $M\circ N$ considered as a bilinear form acts on vectors $a,b$ as

$a^{\ast }(M\circ N)b=\mathrm{tr}(M^{\mathsf{\text{𝖳}}}\mathrm{diag}\left(a^{\ast }\right)N\mathrm{diag}(b))$

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

Suppose $M$ and $N$ are positive definite, and so Hermitian. We can consider their square-roots $M^{\frac{1}{2}}$ and $N^{\frac{1}{2}}$, which are also Hermitian, and write

$\mathrm{tr}(M^{\mathsf{\text{𝖳}}}\mathrm{diag}\left(a^{\ast }\right)N\mathrm{diag}(b))=\mathrm{tr}({\bar{M}}^{\frac{1}{2}}{\bar{M}}^{\frac{1}{2}}\mathrm{diag}\left(a^{\ast }\right)N^{\frac{1}{2}}{N}^{\frac{1}{2}}\mathrm{diag}(b))=\mathrm{tr}({\bar{M}}^{\frac{1}{2}}\mathrm{diag}\left(a^{\ast }\right)N^{\frac{1}{2}}{N}^{\frac{1}{2}}\mathrm{diag}(b){\bar{M}}^{\frac{1}{2}})$

Then, for $a=b$, this is written as $\mathrm{tr}\left(A^{\ast }A\right)$ for $A=N^{\frac{1}{2}}\mathrm{diag}(a){\bar{M}}^{\frac{1}{2}}$ and thus is strictly positive for $A\ne 0$, which occurs if and only if $a\ne 0$. This shows that $(M\circ N)$ is a positive definite matrix.

Proof using Gaussian integration

Case of M = N

Let $X$ be an $n$-dimensional centered Gaussian random variable with covariance $⟨X_i{X}_j⟩=M_{ij}$. Then the covariance matrix of $X_i^2$ and $X_j^2$ is

$\mathrm{Cov}(X_i^2,X_j^2)=⟨X_i^2{X}_j^2⟩-⟨X_i^2⟩⟨X_j^2⟩$

Using Wick's theorem to develop $⟨X_i^2{X}_j^2⟩=2{⟨X_i{X}_j⟩}^2+⟨X_i^2⟩⟨X_j^2⟩$ we have

$\mathrm{Cov}(X_i^2,X_j^2)=2{⟨X_i{X}_j⟩}^2=2M_{ij}^2$

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

General case

Let $X$ and $Y$ be $n$-dimensional centered Gaussian random variables with covariances $⟨X_i{X}_j⟩=M_{ij}$, $⟨Y_i{Y}_j⟩=N_{ij}$ and independent from each other so that we have

$⟨X_i{Y}_j⟩=0$ for any $i,j$

Then the covariance matrix of $X_i{Y}_i$ and $X_j{Y}_j$ is

$\mathrm{Cov}(X_i{Y}_i,X_j{Y}_j)=⟨X_i{Y}_i{X}_j{Y}_j⟩-⟨X_i{Y}_i⟩⟨X_j{Y}_j⟩$

Using Wick's theorem to develop

$⟨X_i{Y}_i{X}_j{Y}_j⟩=⟨X_i{X}_j⟩⟨Y_i{Y}_j⟩+⟨X_i{Y}_i⟩⟨X_j{Y}_j⟩+⟨X_i{Y}_j⟩⟨X_j{Y}_i⟩$

and also using the independence of $X$ and $Y$, we have

$\mathrm{Cov}(X_i{Y}_i,X_j{Y}_j)=⟨X_i{X}_j⟩⟨Y_i{Y}_j⟩=M_{ij}{N}_{ij}$

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

Proof using eigendecomposition

Proof of positive semidefiniteness

Let $M=\sum {\mu }_i{m}_i{m}_i^{\mathsf{\text{𝖳}}}$ and $N=\sum {\nu }_i{n}_i{n}_i^{\mathsf{\text{𝖳}}}$. Then

$M\circ N=\sum _{ij}{\mu }_i{\nu }_j\left(m_i{m}_i^{\mathsf{\text{𝖳}}}\right)\circ \left(n_j{n}_j^{\mathsf{\text{𝖳}}}\right)=\sum _{ij}{\mu }_i{\nu }_j(m_i\circ n_j){(m_i\circ n_j)}^{\mathsf{\text{𝖳}}}$

Each $(m_i\circ n_j){(m_i\circ n_j)}^{\mathsf{\text{𝖳}}}$ is positive semidefinite (but, except in the 1-dimensional case, not positive definite, since they are rank 1 matrices). Also, ${\mu }_i{\nu }_j>0$ thus the sum $M\circ N$ 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 $a\ne 0$, we have $a^{\mathsf{\text{𝖳}}}(M\circ N)a>0$. Continuing as above, each $a^{\mathsf{\text{𝖳}}}(m_i\circ n_j){(m_i\circ n_j)}^{\mathsf{\text{𝖳}}}a\ge 0$, so it remains to show that there exist $i$ and $j$ for which corresponding term above is nonzero. For this we observe that

$a^{\mathsf{\text{𝖳}}}(m_i\circ n_j)(m_i\circ n_j{)}^{\mathsf{\text{𝖳}}}a={\left(\sum _k{m}_{i,k}{n}_{j,k}{a}_k\right)}^2$

Since $N$ is positive definite, there is a $j$ for which $n_j\circ a\ne 0$ (since otherwise $n_j^{\mathsf{\text{𝖳}}}a=\sum _k(n_j\circ a{)}_k=0$ for all $j$), and likewise since $M$ is positive definite there exists an $i$ for which $\sum _k{m}_{i,k}(n_j\circ a{)}_k=m_i^{\mathsf{\text{𝖳}}}(n_j\circ a)\ne 0.$ However, this last sum is just $\sum _k{m}_{i,k}{n}_{j,k}{a}_k$. Thus its square is positive. This completes the proof.

External links

• Bemerkungen zur Theorie der beschränkten Bilinearformen mit unendlich vielen Veränderlichen at EUDML