Product of 2 positive definite matrix 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,
 A: 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.
A: 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).
