Projector has Norm 1
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Theorem
An idempotent operator $P$ is a projector on the Hilbert Space $H$ if and only if $P$ has norm $1$:
- $\ds \norm P \equiv \sup_{x \mathop \in H} \frac {\norm P} {\norm x} = 1$
Proof
For all $x \in \Rng P$:
- $\norm {P \dfrac x {\norm x} } = \dfrac {\norm x} {\norm x} = 1$
so $\norm P \ge 1$.
It remains to show that this holds with equality if and only if $P$ is a projector.
First, suppose $P$ is a projector.
Let $\set {p_1, p_2, \ldots}$ be an orthonormal basis for $\Rng P$.
Let $\set {q_1, q_2, \ldots}$ be an orthonormal basis for $\Rng P_\perp$.
Then for any $x \in H$, we can choose scalars $\set {\alpha_1, \alpha_2, \ldots}$ and $\set {\beta_1, \beta_2, \ldots}$ so that:
- $\ds x = \sum_{i \mathop = 1}^\infty \alpha_i p_i + \sum_{i \mathop = 1}^\infty \beta_i q_i$
Because the basis vectors are orthogonal, Pythagoras's theorem shows that:
- $\ds \norm x^2 = \sum_{i \mathop = 1}^\infty \norm {\alpha_i}^2 + \norm {\beta_i}^2$
Then $\norm {P x}$ can be expanded thus:
\(\ds \norm {P x}^2\) | \(=\) | \(\ds \norm {\sum_{i \mathop = 1}^\infty \alpha_i P p_i + \sum_{i \mathop = 1}^\infty \beta_i P q_i}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\sum_{i \mathop = 1}^\infty \alpha_i p_i + 0}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^\infty \size {\alpha_i}^2\) | Pythagoras's theorem again | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm x^2\) |
Hence $\norm P \le 1$.
Since it was already shown $\norm P \ge 1$, it follows that $\norm P = 1$.
Now suppose $P$ is not a projector.
Then there exists $x \in H$ so that $P x - x$ is not orthogonal to $\Rng P$.
By writing $x = p + q$ with $p \in \Rng P$ and $q \in \Rng P_\perp$, it follows that:
- $P x - x = P \paren {p + q} - \paren {p + q} = P q - q$
By rescaling $x$, we can assume $\norm q = 1$.
Since $q \in \Rng P_\perp$ but
- $P q - q = P x - x = \notin \Rng P_\perp$
it follows that $P q \ne 0$.
Let $\norm {P q} = c \ne 0$.
It will be shown that
- $y = c q + \dfrac 1 c P q$
satisfies
- $\dfrac {\norm {P y} } {\norm y} > 1$
Notice first that:
\(\ds \norm y^2\) | \(=\) | \(\ds \norm {c q + \dfrac 1 c P q}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {c q}^2 + \norm {\dfrac 1 c P q}^2\) | Pythagoras's Theorem ($q$ and $P q$ are orthogonal) | |||||||||||
\(\ds \) | \(=\) | \(\ds c^2 + 1\) | since $\norm {P q} = c$ and $\norm q = 1$. |
Now:
\(\ds \norm {P \frac y {\norm y} }^2\) | \(=\) | \(\ds \frac 1 {\norm y^2} \norm {P \paren {c q + \frac 1 c P q} }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {c^2 + 1} \norm {c P q + \frac 1 c P^2 q}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {c^2 + 1} \norm {P q \paren {c + \frac 1 c} }^2\) | $P$ is idempotent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm {P q}^2} {c^2 + 1} \paren {c + \frac 1 c}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {c^2 \paren {c + \frac 1 c}^2} {c^2 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c^2 + 1\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 1\) |
$\blacksquare$