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.
 A: It's not clear to me what you mean by matrix inequality, but assuming you mean an inequality componentwise, it's clearly false, for instance let
$$A = B = \frac{1}{2} \left[\begin{array}{cc}1 & 1\\1 &1\end{array}\right]$$
be projection onto the vector $(1,1)$.
Then $AB = A$ has positive off-diagonal entries so violates your inequality.
EDIT: $(I-AB)$ is not positive semi-definite, as projection matrices are not in general symmetric. For instance
$$A = \left[\begin{array}{cc}1 & 2\\0 & 0\end{array}\right]$$
is a projection matrix ($A^2 = A$) but $I - A^2 = I-A$ is clearly not symmetric.
A: 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
$\mathbf{x}^T(I-AB)\mathbf{x} \geq 0$.
