# Projector has Norm 1

## Theorem

An idempotent operator $P$ is a projector on the Hilbert Space $H$ if and only if $P$ has norm $1$:

- $\displaystyle \norm P \equiv \sup_{x \mathop \in H} \frac {\norm P} {\norm x} = 1$

## 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 write choose scalars $\set {\alpha_1, \alpha_2, \ldots}$ and $\set {\beta_1, \beta_2, \ldots}$ so that:

- $\displaystyle x = \sum_{i \mathop = 1}^\infty \alpha_i p_i + \sum_{i \mathop = 1}^\infty \beta_i q_i$

Because the basis vectors are orthogonal, the Pythagorean theorem shows that:

- $\displaystyle \norm x^2 = \sum_{i \mathop = 1}^\infty \norm {\alpha_i}^2 + \norm {\beta_i}^2$

Then $\norm {P x}$ can be expanded thus:

\(\displaystyle \norm {P x}^2\) | \(=\) | \(\displaystyle \norm {\sum_{i \mathop = 1}^\infty \alpha_i P p_i + \sum_{i \mathop = 1}^\infty \beta_i P q_i}^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \norm {\sum_{i \mathop = 1}^\infty \alpha_i p_i + 0}^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^\infty \size {\alpha_i}^2\) | by the Pythagorean theorem again | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \norm x^2\) |

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

\(\displaystyle \norm y^2\) | \(=\) | \(\displaystyle \norm {c q + \dfrac 1 c P q}^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \norm {c q}^2 + \norm {\dfrac 1 c P q}\^2\) | by the Pythagorean theorem ($q$ and $P q$ are orthogonal) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle c^2 + 1\) | since $\norm {P q} = c$ and $\norm q = 1$. |

Now

\(\displaystyle \norm {P \frac y {\norm y} }^2\) | \(=\) | \(\displaystyle \frac 1 {\norm y^2} \norm {P \paren {c q + \frac 1 c P q} }^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {c^2 + 1} \norm {c P q + \frac 1 c P^2 q}^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {c^2 + 1} \norm {P q \paren {c + \frac 1 c} }^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\norm {P q}^2} {c^2 + 1} \paren {c + \frac 1 c}^2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {c^2 \paren {c + \frac 1 c}^2} {c^2 + 1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle c^2 + 1\) | |||||||||||

\(\displaystyle \) | \(>\) | \(\displaystyle 1\) |

$\blacksquare$