Does the product of positive definite matrices have a positive trace Let $ A_{1}, A_{2}, \ldots, A_{m}$ be a real symmetric semi-positive  definite matrices,  I want to know whether 
$ tr (A_{1} \cdot A_{2} \cdots A_{m} ) \geq 0$ ?
When $m=2$, it seems a rather standard problem and has a "yes" answer.
 A: For $m\ge3$, no. Counterexample:
$$
A = \pmatrix{6&-2\\ -2&1},
\ B = \pmatrix{1&-2\\ -2&8},
\ C = \pmatrix{10&8\\ 8&10},
\ ABC = \pmatrix{-124&-200\\ 56&88}.
$$
Note that $A,B,C$ are positive definite.
A: Take $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. We have 
$$
\begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}
 = \begin{bmatrix} b & -a \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} 
 = 0 \ge 0.
$$
(Of course, $Ax$ is the rotation of $x$ by $\pi/2$, so $Ax \cdot x = 0$ always.) So $A$ is semi-positive definite. But
$$
A^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
$$
which has negative trace.
So I'm saying your assertion is false because I have a counter example.
EDIT : Just so that I don't look stupid, the symmetric assumption was not written in the question at the time this answer was written.
Hope that helps,
A: The general case should not be true for $m \ge 3$. Look at the proof for $m=2$:
$$  \DeclareMathOperator{\tr}{trace}
    \tr(AB) = \tr(A^{1/2} B A^{1/2} )
$$
which is clearly non-negative since the last matrix is symmetric positive definite.  But this proof does not work for $m \ge 3$ (try it! and you will see) so we should expect an counterexample in that case.  You should simple program a search for counterexamples in the three-matrix case, you will succeed.
