This should be easy for you.
I have an intuitive feeling for why the following should be correct, but I would like something more rigorous than my feeling :)
Consider the $m\times m$-matrix $B$, which is symmetric and positive definite (full rank). Now this matrix is transformed using another matrix, say $A$, in the following manner: $A B A^T$. The matrix $A$ is $n\times m$ with $n<m$. Furthermore the constraint $rank(A) < n$ is imposed.
My intuition tells me that $A B A^T$ must be symmetric and positive semi-definite, but what is the mathematical proof for this? (why exactly does the transformation preserve symmetry and why is it that possibly negative eigenvalues in $A$ still result in the transformation to be PSD? Or is my intuition wrong)?
Edit: please exclude the case of A=0.
For symmetry: note that in general, we have $(AB)^T = B^TA^T$, hence $(ABC)^T = C^TB^TA^T$. With that, we see that $$ (ABA^T)^T = A^{TT}B^TA^T = ABA^T. $$ For positive semidefiniteness: an $n \times n$ symmetric matrix $M$ is positive semidefinite if (and only if) $x^TMx \geq 0$ whenever $x \in \Bbb R^n$. We note that $$ x^T(ABA^T)x = (x^TA)B(A^Tx) = (A^Tx)^T B (A^Tx). $$ Because $B$ is positive definite (and hence positive semidefinite), we must have $y^TBy \geq 0$ for $y = A^Tx$. Thus, $x^T(ABA^T)x \geq 0$, so that $ABA^T$ is indeed positive semidefinite.