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I got a problem I can't solve for in linear algebra. My task is to find if the product of 2 positive definite matrices is also positive definite? My intuition tells me it is not true but I cannot find a counterexample.

If my intuition was false, do this 2 matrices need to be symmetric to be true?

Thanks a lot for your help,

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  • $\begingroup$ Use the property that a matrix is positive definite iff it's eigenvalues are positive. A positive definite matrix is symmetric by definition. $\endgroup$ – Jules Jan 30 '18 at 8:53
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    $\begingroup$ @Jules: Thats not correct. A positive definite matrix does not necessarily has to be symmetric. E.g. consider the $2\times 2$ matrix with ones on the diagonal and a one on the $12$ position as well. It is clearly positive definite but certainly not symmetric $\endgroup$ – Konstantin Jan 30 '18 at 9:10
  • $\begingroup$ That's why I said that positive definite matrices are symmetric by definition. Perhaps some authors have a different definition, but this is the most common one. $\endgroup$ – Jules Jan 30 '18 at 13:04
  • $\begingroup$ Otherwise, consider (0 1, -1 0)^2. $\endgroup$ – Jules Jan 30 '18 at 13:20
  • $\begingroup$ Although, if we require symmetry in the definition then it is still false because the product of two symmetric matrices is not necessarily symmetric. $\endgroup$ – Jules Jan 30 '18 at 13:27
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I found an answer on another forum finally. For me, a matrix is positive definite if xTAx>0 for all column vector x.

We have A=[1,2;2,5] and B=[1,-1;-1,2] 2 positive definite matrix

AB=[-1,3;-3,8] is not positive definite as [1,0]AB[1;0]=-1

The issue with eigenvalue and positive definite matrix is that there are not equivalent -> If the matrix is positive definite then its eigenvalue are positive. But positive eigenvalues does not guarantee that the matrix is positive definite.

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We can define a (not necessarily symmetric) real $n \times n$ matrix $A$ to be positive definite if ${\bf x}^T A {\bf x} \geq 0$ for all ${\bf x} \in \mathbb{R}^n$ and equality holds if and only if $x={\bf 0}$.

Then $$A = \begin{pmatrix} 1& 2a\\ 0& 1 \end{pmatrix}$$ is positive definite if and only if $|a|<1$. And so is the transpose. But, if you try to compute the product of some of these matrices, you'll find a product to two which is no longer definite.

Here's a non-symmetric positive definite matrix which squares to a indefinite matrix: $$L = \begin{pmatrix} 1& 1\\ -1& 1 \end{pmatrix}.$$

For symmetric positive definite matrices, as Jules says in the comments above, the product of two symmetric matrices need not be symmetric. Indeed, if $A$, $B$, and $AB$ are symmetric, then we must have that $AB = (AB)^T = B^T A^T = BA$. However, since every matrix commutes with itself, the square of a positive definite matrix is positive definite.

This note may be helpful regarding whether or not to insist that the matrix be defined as a symmetric matrix. The example below is inspired by this note.

A search of "non symmetric positive definite matrix" will yield this and many other discussions including some here on math.stackexchange.

One final example: consider $q(x,y) = {\bf x}^T A {\bf x} = x^2 + 2axy + y^2$. We find that the symmetric matrix associated to this quadratic form is

$$M = \begin{pmatrix} 1& a\\ a& 1 \end{pmatrix},$$ so that $q(x,y) = {\bf x}^T M {\bf x}$. This is positive definite if and only if its determinant is positive (which proves the claim about the matrix $A$ above).

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