Schur product theorem

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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[1] (Schur 1911, p. 14, Theorem VII) (note that Schur signed as J. Schur in Journal für die reine und angewandte Mathematik.[2][3])

Proof[edit]

Proof using the trace formula[edit]

It is easy to show that for 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 .

Since and are positive definite, we can consider their square-roots and and write

Then, for , this is written as for and thus is positive. This shows that is a positive definite matrix.

Proof using Gaussian integration[edit]

Case of M = N[edit]

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[edit]

Let and be -dimensional centered Gaussian random variables with covariances , and independent from each other so that we have

for any

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[edit]

Proof of positive semidefiniteness[edit]

Let and . Then

Each is positive semidefinite (but, except in the 1-dimensional case, not positive definite, since they are rank 1 matrices). Also, thus the sum is also positive semidefinite.

Proof of definiteness[edit]

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 the inequality is strict. For this we observe that

Since is positive definite, there is a for which is not 0 for all , and then, since is positive definite, there is an for which is not 0 for all . Then for this and we have . This completes the proof.

References[edit]

  1. ^ "Bemerkungen zur Theorie der beschränkten Bilinearformen mit unendlich vielen Veränderlichen". Journal für die reine und angewandte Mathematik (Crelle's Journal) 1911 (140): 1–00. 1911. doi:10.1515/crll.1911.140.1. 
  2. ^ Zhang, Fuzhen, ed. (2005). "The Schur Complement and Its Applications". Numerical Methods and Algorithms 4. doi:10.1007/b105056. ISBN 0-387-24271-6. , page 9, Ch. 0.6 Publication under J. Schur
  3. ^ Ledermann, W. (1983). "Issai Schur and His School in Berlin". Bulletin of the London Mathematical Society 15 (2): 97–106. doi:10.1112/blms/15.2.97. 

External links[edit]