# Idempotent Operators

For a finite-dimensional Hilbert space $H$, a linear operator $P:H\to H$ is called "idempotent" if $P^2=P$ ($Px=x$ for $x\in range(P)$ is an immediate consequence). An idempotent operator is a "projector" if $\forall x\in H:Px-x\perp range(P)$. Some texts refer to these as "projectors" and "orthogonal projectors" respectively.

## Theorem

All linear idempotent operators $P$ have $\|P\|_2\equiv\sup_{x\in X}\|Px\|_2\geq1$ with equality iff $P$ is a projector.

## Proof

Let $P$ be a linear idempotent operator on $X$. Let $\{q_1,\dots,q_m\}$ be an orthonormal basis for $PX$ and $\{q_{m+1},\dots,q_n\}$ be an orthonormal basis for $(PX)_\perp$.

For any $x\in PX=range(P)$: $\|P\frac{x}{\|x\|}\|=\frac{\|x\|}{\|x\|}=1$ so $\|P\|\geq1$.

Suppose $P$ is a projector. Then for any $\|\sum_{i=1}^n q_i\|=1$, we have $\|P\sum_{i=1}^n q_i\|=\|\sum_{i=1}^ n Pq_i\|=\|\sum_{i=1}^ m q_i\|\leq1$, hence $\|P\|\leq1$ Since we have already shown $\|P\|\geq1$, $\|P\|=1$.

Suppose $P$ is not a projector. Then we may choose $q=q_j,j>m$ so that $\|Pq\|=c\neq0$. We claim that $x=q\|Pq\|+Pq/\|Pq\|$ satisfies $P\frac{x}{\|x\|}>1$.

Notice first $\|x\|^2=\|qc+Pq/c\|^2=\|qc\|^2+\|Pq/c\|^2$ since $q$ and $Pq$ are orthogonal. Now

\begin{align} \left\|P\frac{x}{\|x\|}\right\|^2 =&\|P(q\|Pq\|+Pq/\|Pq\|)\|^2/\|x\|^2\\ =&\left\|Pqc+P^2q/c\right\|^2/\|x\|^2\\ =&\left\|Pq(c+1/c)\right\|^2/\|x\|^2\\ =&\|Pq\|^2(c+1/c)^2/\|x\|^2\\ =&\frac{c^2(c+1/c)^2}{\|x\|^2}\\ =&\frac{c^2(c+1/c)^2}{\|qc\|^2+\|Pq/c\|^2}\\ =&\frac{c^2(c+1/c)^2}{\|q\|^2c^2+\|Pq\|^2/c^2}\\ =&\frac{c^2(c+1/c)^2}{c^2+1}\\ =&c^2+1\\ >&1 \end{align}