Non-commuting orthogonal projections I have some bad intuition. Can anyone provide me with two concrete non-commuting orthogonal projections on a Hilbert space?
 A: Consider $\mathbb{R}^2$ with the standard scalar product and consider the two subspaces $$S=\operatorname{span}(\begin{pmatrix} 1 \\ 0 \end{pmatrix})\quad \text{and} \quad T= \operatorname{span}(\begin{pmatrix} 1 \\ 1 \end{pmatrix}).$$
Let $P_S$ and $P_T$ denote the projection mappings onto $S$ and $T$ respectively and  verify that
$$P_T(P_S(\begin{pmatrix} 0 \\ 1 \end{pmatrix})) = P_T(\begin{pmatrix} 0 \\ 0 \end{pmatrix})= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
while
$$P_S(P_T(\begin{pmatrix} 0 \\ 1 \end{pmatrix})) = P_S(\begin{pmatrix} \frac{1}{2} \\ \frac{1}{2} \end{pmatrix})= \begin{pmatrix} \frac{1}{2} \\ 0 \end{pmatrix}.$$
So the projection mappings do not commute, not even in the finite dimensional case.
A: @Leander's answer is perfect, but it might be interesting to see the concrete matrices associated with the two projections given:
$$
P_S = \pmatrix{1 & 0 \cr 0 & 0},
\qquad
P_T= \pmatrix{{1\over 2} & {1\over 2} \cr {1\over 2} & {1\over 2}},
$$
so that one may verify algebraically that they do not commute:
$$
P_SP_T = \pmatrix{{1\over 2} & {1\over 2} \cr 0 & 0},
$$
while
$$
P_TP_S = \pmatrix{{1\over 2} & 0 \cr {1\over 2} & 0}.
$$
