Lemma 3

Let E be a Hilbert space and $p\in \mathscr{L}(E)$ a projector. The following assertions are equivalent:

(i) The projector $p$ is an orthogonal projector, that is to say Ker($p$) $=$ Im($p$)$^{\circ}$ (EVT, V, p. 13);

(ii) The projector $p$ is hermitian;

(iii) The projector $p$ is normal;

(iv) One has Ker($p$)$\subset$ Ker($p^*$) ;

(v) One has Im($p$)$\subset$ Im($p^*$) ;

(vi) The projector $p$ is positive;

(vii) One has $\|p\|\leqslant 1$.

Let us first recall that Ker($p^*$) $=$ Im($p$)$^{\circ}$ and Ker($p$) $=$ Im($p^*$)$^{\circ}$ (EVT, V, p. 41, prop. 4). Moreover, the image of $p$ (resp. $p^*$) is closed, since it coincides with the kernel of $1-p$ (resp. $1-p^*$). Therefore one has

(1) Im($p$) $=$ Ker($p^*$)$^{\circ}$, Im($p^*$) $=$ Ker($p$)$^{\circ}$.

(i) $=\Rightarrow$ (ii)$:p^*$ is a projector with kernel Im($p$)$^{\circ}=$ Ker($p$) and whose image is Im($p^*$)$^{\circ}=$ Ker($p$) $=$ Im($p$)$^{\circ}$; hence $p^*=p$.

(ii) $=\Rightarrow$ (iii) since every hermitian endomorphism is normal.

(iii) $=\Rightarrow$ (iv) : since $p$ is normal, one has $\|p(x)\|^2=\|p^*(x)\|^2$ for all $x$ in E (EVT, V, p. 43, Prop. 7), whence the required inclusion.

(iv) $=\Rightarrow$ (v) follows from equalities (1) above.

(v) $=\Rightarrow$ (vi) : for all $x\in E$, one has $p(x)\in$ Im($p^*$), and consequently $\langle p(x)|x\rangle =\langle p^*(p(x))|x\rangle =\|p(x)\|^2\geqslant 0$.

(vi) $=\Rightarrow$ (vii) : let $x\in E$ and $y=x-p(x)\in$ Ker($p$). For all $t\in \mathbf{R}$, one has by hypothesis

which is possible only if $\langle y|p(x)\rangle = 0$. But then

and therefore $\|p\|\leqslant 1$.

(vii) $=\Rightarrow$ (i) : let $y\in$ Im($p$) ; denote by $z$ the orthogonal projection of $y$ onto Ker($p$)$^{\circ}$ and put $x=y-z\in$ Ker($p$). One has $p(z) =p(y) =y$, hence $\|y\|\leqslant \|z\|$ by hypothesis. But, since $x$ and $z$ are orthogonal, one has $\|y\|^2=\|x\|^2+\|z\|^2$, whence $\|x\|= 0$, that is to say $y=z$. Thus, Im($p$)$\subset$ Ker($p$)$^{\circ}$. Since moreover $\|p^*\|=\|p\|\leqslant 1$, one has analogously Im($p^*$)$\subset$ Ker($p^*$)$^{\circ}$, which yields the converse inclusion by (1).