# Test of whether matrix is Symmetric Positive Definite is giving wrong result when matrix is not symmetric

14 views (last 30 days)

Show older comments

I'm computing a multivariate normal probability density with an estimated covariance matrix as follows:

% Update the conditional likelihood given the data

p_yk_g_seq_Ykm1(j) = mvnpdf(yk, ykp1_est, Sk);

and getting the following error

Error using mvnpdf (line 127)

SIGMA must be a square, symmetric, positive definite matrix.

Here are some checks I did in the debugger when this error occurred:

K>> Sk

Sk =

0.0540 -0.0001

-0.0001 0.0540

K>> issymmetric(Sk)

ans =

logical

0

Clearly it is not symmetric.

But when I tried to check if Sk is symmetic positive definite using the method described here in the documentaion:

K>> try chol(Sk)

disp('Matrix is symmetric positive definite.')

catch ME

disp('Matrix is not symmetric positive definite')

end

ans =

0.2323 -0.0002

0 0.2323

Matrix is symmetric positive definite.

This confused me. What is going on here? Is it non-symmetric or not PD or both?

Looking in the code for mvnpdf it actually uses this check:

[R,err] = cholcov(Sigma,0);

So maybe the method from the documentation page linked above is wrong or out of date?

Or maybe it tests for positive semi-definite but not symmetric positive semi-definite.

##### 0 Comments

### Accepted Answer

Steven Lord
on 8 Nov 2022

From the chol documentation page: "If A is nonsymmetric , then chol treats the matrix as symmetric and uses only the diagonal and upper triangle of A." Is the symmetric matrix generated using the diagonal and upper triangle of your matrix SPD?

I'll ask the documentation staff to take a look at the page to which you linked.

##### 4 Comments

Matt J
on 9 Nov 2022

### More Answers (1)

Matt J
on 8 Nov 2022

Edited: Matt J
on 9 Nov 2022

But when I tried to check if Sk is symmetic positive definite using the method described here in the documentaion:

All of the tests at this link are predicated on the assumption that Sk is already symmetric. In other words, they are really intended as tests of postive or non-negative definiteness, not of symmetry.

Therefore, you should just symmetrize the matrix and be done with it,

Sk=(Sk+Sk.')/2;

This does not change the status of Sk as a PSD or PD matrix.

##### 3 Comments

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!