Dear Ashraf, product of posdef matrix A and negdef matrix B is a negdef matrix:

if uTAu > 0 for every u, uTBu < 0 for every u, then uTABu = uTAu uTBu/(uuT) < 0 because uuT > 0 .

Gianluca

Dear Ashraf, product of posdef matrix A and negdef matrix B is a negdef matrix:

if uTAu > 0 for every u, uTBu < 0 for every u, then uTABu = uTAu uTBu/(uuT) < 0 because uuT > 0 .

Gianluca

Dear Ashraf,

No, this is not the case. Consider the counter example:

A = I

B = -I

in which I is the identity matrix. In this case A is positive definite and B is negative definite with their product be negative definite.

Mirko

First, notice that the product is not necessarily symmetric, except if the matrices commute,

Suppose M and N two symmetric positive-definite matrices and λ ian eigenvalue of the product MN. Then it's possible to show that  λ>0 and thus MN has positive eigenvalues. Now, take M symmetric positive-definite and N symmetric negative-definite. As a result, apply the previous result to -(MN) then MN have negative eigenvalues. I hope this could be fairly clear,

Fabrizio

Dear Fabrizio, Mirko and Gianluca, thank you very much your answers were very helpful.

Best Regards

Ashraf

Remember: positive or negative-definite is not a matrix property but it only applies to quadratic forms, which are naturally described only by symmetric matrices.

For instance, a way to establish positive definiteness of a quadratic form is to find this symmetric matrix representing it and test whether its eigenvalues are all positive.

But there exists infinitely many matrices representing a particular quadratic form, all with and exactly one of them is symmetric. Thus it's possible to have non-symmetric definite matrices.

Furthermore, it could be showed that for a not necessarily symmetric matrix to be

positive definite it's necessary but not sufficient that its real eigenvalues are all positive.

A small observation:

There are good answers, yet, to complete Fabrizio’s answer, the symmetry in positive definite matrices is a property with which we got used only because it appears in many examples.

However, symmetry is NOT needed for a matrix to be positive definite.

As people mentioned, the property comes from the quadratic form, which is defined to be positive definite, namely, the scalar product r=x'Mx>0 for any vector x≠0.

Because the result r is scalar, we clearly have r=r'.

Therefore, even if M is not symmetric, we may still have r=x'Mx=x'M'x >0. When M is symmetric, this is clear, yet iin general, it may also happen if M≠M'.

Of course, if the nonsymmetric matrix M is positive definite, so is its symmetric component Ms=( M+M')/2.

Dear Fabrizio and Itzhak thank you for the valuable contributions.

Dear Gianluca

Notice that $uu^T$ is not a scaler. It is a square matrix, therefore your proof is not true.