Non-commuting orthogonal projections

Question

I have some bad intuition. Can anyone provide me with two concrete non-commuting orthogonal projections on a Hilbert space?

Answer 1

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.

Answer 2

@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 $$ P_TP_S = \pmatrix{{1\over 2} & 0 \cr {1\over 2} & 0}. $$