Proposition 9

— Let E be a hilbertian space and M, N two closed vector subspaces of E.

a) The following conditions are equivalent : (i) $p_M p_N = p_N p_M$; (ii) if $x \in M$ is orthogonal to $M \cap N$ and if $y \in N$ is orthogonal to $M \cap N$, then x and y are orthogonal; (iii) every vector of M orthogonal to $M \cap N$ is orthogonal to N; (iv) $M = (M \cap N) + (M \cap N^\circ)$.

b) If the equivalent conditions of a) are satisfied, we have $p_{M \cap N} = p_M p_N$, the vector subspace $M + N$ of E is closed and we have $p_{M+N} = p_M + p_N - p_M p_N$.

c) We have $p_M p_N = 0$ if and only if M is orthogonal to N. If this is so, then the vector subspace $M + N$ of E is closed, and $p_{M+N} = p_M + p_N$.

Put $L = M \cap N$, $M_1 = M \cap L^\circ$ and $N_1 = N \cap L^\circ$. Condition (ii) implies that $M_1$ and $N_1$ are orthogonal, and (iii) implies that $M_1$ and N are orthogonal. Since we have $N = N_1 + L$ and $M_1$ is orthogonal to L, we have proved the equivalence of (ii) and (iii). If condition (iii) is satisfied, we have $M_1 = M \cap N^\circ$ and since $M = L + M_1$, condition (iv) is satisfied. Conversely, from (iv) we conclude that $M_1 = M \cap N^\circ$ since the subspaces $M \cap N$ and $M \cap N^\circ$ of M are orthogonal, and so $M_1 \subset N^\circ$, that is, the relation (iii).

Assume that condition (iv) is satisfied. It is immediate that $p_N(y) = p_L(y)$ for all $y \in M$ and hence $p_N p_M(x) = p_L p_M(x)$ for all $x \in E$. But, for every $x \in E$, the vector $p_L p_M(x)$ belongs to L, and the vector

belongs to $M^\circ + L^\circ = L^\circ$; hence we have $p_L p_M(x) = p_L(x)$. Finally, $p_N p_M = p_L p_M = p_L$. Since condition (ii) is equivalent to (iv) and is symmetric in M and N, we also have $p_M p_N = p_L$. Finally we get $p_M p_N = p_N p_M = p_{M \cap N}$ which gives (i).

Conversely, suppose condition (i) is satisfied. Let $x \in M$; we have

and so $p_N(x) \in M$. We conclude that $x - p_N(x) \in M$, hence $x$ is the sum of an element $p_N(x)$ of $M \cap N$ and an element $x - p_N(x)$ of $M \cap N^\circ$, which gives (iv).

We have proved a) and the first part of b). Assume now that $p_M$ and $p_N$ commute and put $q = p_M + p_N - p_M p_N$; since $p_M$ and $p_N$ are idempotents in the algebra $\mathscr{L}(E)$, so is q; hence (GT, III, § 6, No. 2) the image of q is a closed vector subspace of E.

It is clear that the image of $q$ is contained in $M + N$; however, we have $p_N(x) = x$, hence $q(x) = x$ for all $x \in N$; since we also have $q = p_M + p_N - p_N p_M$, we get $q(x) = x$ for all $x \in M$. We conclude that the image of $q$ is equal to $M + N$. The orthogonal of $M + N$ is equal to $M^\circ \cap N^\circ$, and the kernel of $q$ obviously contains $M^\circ \cap N^\circ$, hence $q = p_{M+N}$. This proves b).

We have $p_M p_N = 0$ if and only if the image $N$ of $p_N$ is contained in the kernel $M^\circ$ of $p_M$, that is, if and only if $M$ is orthogonal to $N$. The rest of the assertion c) is then a particular case of b).