Projection/Idempotent Linear Operators

# Projection/Idempotent Linear Operators

Definition: Let $X$ be a linear space. A linear operator $P : X \to X$ is said to be a Projection or Idempotent if $P^2 = P$, that is, $P(P(x)) = P(x)$ for every $x \in X$. |

If $X$ is a linear space and $P : X \to X$ is a projection, then the kernel of $P$ has a nice form which is proven in the lemma below.

Lemma 1: Let $X$ be a linear space and let $P : X \to X$ be a projection. Then $(I - P)(X) = \ker (P)$. |

**Proof:**Let $x \in (I - P)(X)$. Then there exists a $y \in X$ such that:

\begin{align} \quad x = (I - P)(y) = I(y) - P(y) = y - P(y) \end{align}

- Since $P$ is a projection, $P^2 = P$ and thus:

\begin{align} \quad P(x) = P(y - P(y)) =P(y) - P^2(y) = P(y) - P(y) = 0 \end{align}

- Hence $x \in \ker (P)$ and so $(I - P)(X) \subseteq \ker (P)$.

- $\Leftarrow$ Suppose that $x \in \ker (P)$. Then $P(x) = 0$. So then:

\begin{align} \quad x = I(x) = I(x) - 0 = I(x) - P(x) = (I - P)(x) \end{align}

- Hence $x \in (I - P)(X)$ and so $\ker (P) \subseteq (I - P)(X)$. We conclude that:

\begin{align} \quad (I - P)(X) = \ker (P) \quad \blacksquare \end{align}