Lecture 30 - Practice with Gram-Schmidt

As an example, consider $V = P_{2} (R)$ equipped with:

⟨ f, g ⟩ = \int_{0}^{2} f (x) g (x) d x

Here $‖ f ‖^{2} = \int_{0}^{2} f^{2} (x) d x$ . Let $v_{1} = 1, v_{2} = x, v_{3} = x^{2}$ are our standard basis. Our goal is to find an orthonormal basis for $P_{2} (R)$ .

Actually, the process is pretty much exact to Chapter 6 - Inner Product Spaces#An Example, so check that out if you want a better example. It's very similar to this one.

6.C: Orthogonal Complements

Note

This section is just good to know for the final. We'll cover it again in Linear III.

Consider the plane:

Each plane has a unique $\vec{n}$ associated with it, and vice versa. In linear algebra, we want to abstract this idea even further:

orthogonal complement, $U^{⊥}$

If $U$ is a subset (maybe not a subspace) of $V$ , then the orthogonal complement of $U$ , denoted $U^{⊥}$ , is the set of all vectors in $V$ that are orthogonal to every vector in $U$ :

U^{⊥} = {v \in V : ⟨ v, u ⟩ = 0 \forall u \in U}

So $U^{⊥}$ is all the vectors that are orthogonal to the plane, or are on the $\vec{n}$ line:

Some properties include:

$U^{⊥}$ is a subspace of $V$ , even if $U$ is not a subspace but just any subset.
${{\vec{0}}^{⊥}} = V$ , since the zero vector is orthogonal to all $v \in V$ .
$V^{⊥} = {\vec{0}}$ for similar reasoning.
$U \cap U^{⊥} \subseteq {\vec{0}}$ since $\vec{0}$ may or may not be in $U$ , but $U^{⊥}$ always has to have it. All other non-zero vectors are not perpendicular to themselves.
If $U \subseteq W$ then $W^{⊥} \subseteq U^{⊥}$ . (here $W$ is a subset of $V$ , similar to $U$ )

We'll show that it's a subspace (property 1) below:

Proof

Clearly $\vec{0} \in U^{⊥}$ .

Let's show that $U^{⊥}$ is closed under addition. Let $v_{1}, v_{2} \in U^{⊥}$ be arbitrary. So then $⟨ v_{1}, u ⟩ = 0$ and $⟨ v_{2}, 0 ⟩$ for all $u \in U$ . Let $u \in U$ be arbitrary. Notice that:

⟨ v_{1} + v_{2}, u ⟩ = ⟨ v_{1}, u ⟩ + ⟨ v_{2}, u ⟩ = 0 + 0 = 0

Thus since $u$ was arbitrary, then it's clear that $v_{1} + v_{2}$ is perpendicular to all $u$ , so then $v_{1} + v_{2} \in U^{⊥}$ , showing closure under addition.

Closure under scalar multiplication is similar.
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For the proof of property 5 above:

Proof
Suppose $U \subseteq W$ . Let $v \in W^{⊥}$ . Thus then $⟨ v, w ⟩ = 0$ for all $w \in W$ , implying that $⟨ v, u ⟩ = 0$ for every $u \in U$ since $U \subseteq W$ . Hence, $v \in U^{⊥}$ , so then $W^{⊥} \subseteq U^{⊥}$ .
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Some Facts

Lemma

Suppose $U$ is a finite-dimensional subspace of $V$ . Then:

V = U \oplus U^{⊥}

as well as:

(U^{⊥})^{⊥} = U

Proof
To prove the direct sum, clearly $U \cap U^{⊥} = {\vec{0}}$ since the zero vector is in both now ( $U$ is a subspace specifically). Therefore, the sum is direct. We now need to show that $V = U + U^{⊥}$ .

Clearly $U \cap U^{⊥} \subseteq V$ by $U$ being a subspace. Consider the other direction. If $v \in V$ , then let $e_{1}, . . ., e_{k}$ be an orthonormal basis for $U$ . Then:

v = \underset{\in U}{\underset{⏟}{⟨ v, e_{1} ⟩ e_{1} + \dots + ⟨ v, e_{k} ⟩ e_{k}}} + \underset{\in U^{⊥}}{\underset{⏟}{(v - (⟨ v, e_{1} ⟩ e_{1} + \dots + ⟨ v, e_{k} ⟩ e_{k}))}}

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The nice thing here is that the vector on the left-hand side is the "closest" approximation for $v$ in vectors in terms of $V$ .