# Orthogonal Projection Operators Examples 1

Recall from the Orthogonal Projection Operators page that if $U$ is a subspace of $V$, then $V = U \oplus U^{\perp}$ so that for every vector $v \in V$ we have that $v = u + w$ where $u \in U$ and $w \in U^{\perp}$. We defined the orthogonal projection operator $P_U \in \mathcal L(V)$ by $P(v) = P(u + w) = u$ for all $v \in V$.

We also noted the following properties regarding orthogonal projection operators:

- $\mathrm{range} P_U = U$.

- $\mathrm{null} P_U = U^{\perp}$.

- $v - P_U(v) \in U^{\perp}$.

- $P_U^2 = P_U$.

- $\| P_U(v) \| ≤ \| v \|$ for every $v \in V$.

We will now look at some examples regarding orthogonal projection operators.

## Example 1

**Let $V$ be a vector space and let $v_1, v_2, ..., v_m \in V$. Prove that then $\{ v_1, v_2, ..., v_m \}^{\perp} = (\mathrm{span} (v_1, v_2, ..., v_m))^{\perp}$.**

Let $w \in \{ v_1, v_2, ..., v_m \}^{\perp}$. Then we have that $<w, v> = 0$ for every vector $v \in \{ v_1, v_2, ..., v_m \}$. Therefore $<w, v_1> = <w, v_2> = ... = <w, v_n> = 0$. Thus we have that:

(1)So $w$ is orthogonal to every vector in $\mathrm{span} (v_1, v_2, ..., v_m)$. Therefore, $w \in (\mathrm{span} (v_1, v_2, ..., v_m))^{\perp}$.

Now let $w \in (\mathrm{span} (v_1, v_2, ..., v_m))^{\perp}$. Then $w$ is orthogonal to every vector in $\mathrm{span} (v_1, v_2, ..., v_m)$. So $<w, v_1> = 0$, $<w, v_2> = 0$, …, $<w, v_m> = 0$ since $v_1, v_2, ..., v_m \in \mathrm{span} (v_1, v_2, ..., v_m)$. Since $w$ is orthogonal to each of $v_1, v_2, ..., v_m$ we have that $w \in \{ v_1, v_2, ..., v_m \}^{\perp}$.

Therefore $\{ v_1, v_2, ..., v_m \}^{\perp} = (\mathrm{span} (v_1, v_2, ..., v_m))^{\perp}$