Given fat matrices $X$ and $Y$ with full row rank, and diagonal matrix $D$, show that $X^t D Y + Y^t D X$ is never positive definite.
I can see that $X ^t D Y + Y^t D X$ is of the form $A + A^t$. But I am not sure how to proceed. I also know that it suffices to show that there is a negative eigenvalue of this matrix, but can't see how to do that
Following Dan's hints . . .
Let $B$ be an $n{\times}n$ matrix.
Then for any $n{\times}1$ column matrix $x$, we have \begin{align*} x^T(B+B^T)x &=x^TBx+x^TB^Tx\\[4pt] &=x^TBx+(x^TBx)^T\\[4pt] &=x^TBx+x^TBx\qquad\text{[since $x^TBx$ is a $1{\times}1$ matrix]}\\[4pt] &=2(x^TBx)\\[4pt] \end{align*} It follows that $B+B^T$ is positive definite if and only if $B$ is positive definite.
Let $m,n$ be positive integers, with $m < n$.
Suppose $X,Y$ are $m{\times}n$ matrices, and $D$ is an $m{\times}m$ symmetric matrix ($D$ need not be diagonal; symmetric will suffice).
Since $Y^TDX = (X^TDY)^T$, to prove $X^TDY+Y^TDX$ is not positive definite, it suffices to prove $X^TDY$ is not positive definite.
Since $m < n$, there exists a nonzero $n{\times}1$ column matrix $v$ such that $Yv=0$.
But then $v^T(X^TDY)v=0$, so $X^TDY$ is not positive definite.
It follows that $X^TDY+Y^TDX$ is not positive definite.