Schur product theorem


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

Proof

Proof using the trace formula

For any matrices and, the Hadamard product considered as a bilinear form acts on vectors as
where is the matrix trace and is the diagonal matrix having as diagonal entries the elements of.
Suppose and are positive definite, and so Hermitian. We can consider their square-roots and, which are also Hermitian, and write
Then, for, this is written as for and thus is strictly positive for, which occurs if and only if. This shows that is a positive definite matrix.

Proof using Gaussian integration

Case of ''M'' = ''N''

Let be an -dimensional centered Gaussian random variable with covariance. Then the covariance matrix of and is
Using Wick's theorem to develop we have
Since a covariance matrix is positive definite, this proves that the matrix with elements is a positive definite matrix.

General case

Let and be -dimensional centered Gaussian random variables with covariances, and independent from each other so that we have
Then the covariance matrix of and is
Using Wick's theorem to develop
and also using the independence of and, we have
Since a covariance matrix is positive definite, this proves that the matrix with elements is a positive definite matrix.

Proof using eigendecomposition

Proof of positive semidefiniteness

Let and. Then
Each is positive semidefinite. Also, thus the sum is also positive semidefinite.

Proof of definiteness

To show that the result is positive definite requires further proof. We shall show that for any vector, we have. Continuing as above, each, so it remains to show that there exist and for which corresponding term above is non-negative. For this we observe that
Since is positive definite, there is a for which , and likewise since is positive definite there exists an for which However, this last sum is just. Thus its square is positive. This completes the proof.