Multiplication of projection matrices Let $A$ and $B$ be two projection matrices having same dimensions. Then does the following hold
\begin{equation}
AB\leq I,
\end{equation}
where $I$ is the identity matrix. In other words is it true that $I-AB$ is positive semi-definite.

Another answer here shows $AB$ need not be symmetric, so the answer is "no".  But even if you use some nonstandard definition of "positive semidefinite" that doesn't require symmetry, the answer is still no.
Let $n=2$, $A=I$, and $B=\begin{bmatrix}0&1\\0&1\end{bmatrix}$.  Let $\mathbf{x}=\begin{bmatrix}-1\\2\end{bmatrix}$.  Then
$$
\mathbf{x}^T(I-AB)\mathbf{x}=\mathbf{x}^T(I-B)\mathbf{x}=-1<0
$$
But if $A$ and $B$ is are both orthogonal projection matrices ($A^2=A=A^T$ and $B^2=B=B^T$), then, while $AB$ is probably not symmetric, it can be shown that $\|AB\mathbf{x}\| \leq \|\mathbf{x}\|$ for any column vector $\mathbf{x}$ of the right length, where $\|\cdot\|$ is Euclidean norm, so
 $\|\mathbf{x}\|^2 = \mathbf{x}^T \mathbf{x} \geq \|\mathbf{x}\|\|AB\mathbf{x}\|$, and
$\mathbf{x}^T(I-AB)\mathbf{x} \geq 0$.
