Notes - Linear Algebra MT23, Orthogonal complements

Created: November 23, 2023

Flashcards

How does having an inner product $\langle \cdot, \cdot \rangle$ induce a natural injective $\mathbb R$-linear map $\phi : V \to V’$?
\[\phi(v) = \langle v, \cdot \rangle\]

Define the orthogonal complement $U^\perp$ of subspace $U \le V$?
\[U^\perp \triangleq \\{v \in V \mid \langle u, v\rangle = 0, \text{ } \forall u \in U\\}\]

Suppose $U \le V$. In the finite dimensional case, what can you say about $U \oplus U^\perp$?
\[U \oplus U^\perp = V\]

What “dimension formula” do you get with regards to the orthogonal complement $U^\perp$ of a subspace $U \le V$ in the finite dimensional case?
\[\dim V = \dim U + \dim U^\perp\]

For any subspace $U$, what is $U \cap U^\perp$?
\[\\{0\\}\]

How can you rewrite $(U + W)^\perp$?
\[U^\perp \cap W^\perp\]

How can you rewrite $U^\perp \cap W^\perp$?
\[(U + W)^\perp\]

What can you say about the relationship between $(U \cap W)^\perp$ and $U^\perp + W^\perp$?
\[U^\perp + W^\perp \subseteq (U \cap W)^\perp\]

with equality if $\dim V < \infty$.

What can you say about the relationship between $U$ and $(U^\perp)^\perp$?
\[U \subseteq (U^\perp)^\perp\]

with equality if $\dim V < \infty$.

Suppose $V$ is a finite dimensional inner product space. How can you link $U^\perp$ and $U^0$?

Under the natural isomorphism $\phi : V \to V’$ given by $v \mapsto \langle v, \cdot \rangle$, $U^\perp$ maps isomorphically to $U^0$.

Suppose:

  • $V$ is a finite dimensional vector space over $\mathbb R$
  • $\phi : V \to V’$ is the natural $\mathbb R$-linear isomorphism given by $v \mapsto \langle v, \cdot \rangle$

Quickly prove that then $U^\perp$ maps isomorphically to $U^0$.

We use the usual approach of showing one inclusion, checking injectivity, and then using dimensions to check surjectivity.

Suppose $v \in U^\perp$. Then for all $u \in U$:

\[\langle u, v \rangle = \overline{\langle v, u \rangle} = 0\]

Hence $\langle v, \cdot \rangle \in U^0$. Then $\text{Im}(\phi \vert _ {U^\perp}) \subseteq U^0$.

Injective since $\phi(v) = 0$ implies $v = 0$ by non-degeneracy.

Surjective, since

\[\dim U^\perp = \dim V - \dim U = \dim U^0\]

Quickly prove for $U, W \le V$:

  1. $U \cap U^\perp = \{0\}$
  2. $U \oplus U^\perp = V$ if $V$ is finite dimensional
  3. $(U + W)^\perp = U^\perp \cap W^\perp$
  4. $U^\perp + W^\perp \subseteq (U \cap W)^\perp$ with equality if $V$ is finite dimensional
  5. $U \subseteq (U^\perp)^\perp$ with equality if $V$ is finite dimensional

1. $U \cap U^\perp = \{0\}$:

If $u \in U$ and $u \in U^\perp$, then $\langle u, u \rangle = 0$. As $\langle \cdot, \cdot \rangle$ is positive definite, this implies $u = 0$.

2. $U \oplus U^\perp = V$:

Consider an orthonormal basis of $U$ given by $\{e _ 1, \cdots, e _ k\}$ and extend to a basis $\{e _ 1, \cdots, e _ n\}$ of $V$. Consider

\[v = \sum^n_{i = 1} \alpha_i e_i = \sum_{i = 1}^k \alpha_i e_i + \sum^n_{i>k} \alpha_i e_i\]

Then $\sum _ {i = 1}^k \alpha _ i e _ i \in U$ and by taking $\langle e _ \ell, \sum^n _ {i > k} \alpha _ i e _ i\rangle$ for each $1 \le \ell \le k$, we see that $\sum^n _ {i > k} \alpha _ i e _ i \in U^\perp$. Then by part 1, $V = U \oplus U^\perp$.

3. $(U + W)^\perp = U^\perp \cap W^\perp$:

First suppose that $v \in U^\perp \cap W^\perp$. Then $\forall u \in U$ and $\forall w \in W$, \langle v, u \rangle = 0$ and $\langle v, w\rangle = 0$. So $\langle v, u + w \rangle = 0$. Now suppose $v \in (U+W)^\perp$. Then $\forall u, w \in U, W$, $\langle v, u + w\rangle = 0$. By taking $u$ or $w$ as $0$, $\langle v, u\rangle = 0$ and $\langle v, w\rangle = 0$.

4. $U^\perp + W^\perp \subseteq (U \cap W)^\perp$ with equality if $V$ is finite dimensional:

Suppose $v \in U^\perp + W^\perp$. Then $v = u + w$ where $u \in U^\perp$ and $v \in W^\perp$. Hence $\forall s \in U \cap W$, $\langle v, s\rangle = \langle u + w, s \rangle = \langle u, s \rangle + \langle w, s \rangle = 0$. Now suppose we are in finite dimensional case. Using the dimension formula applied to $U^\perp, W^\perp, (U^\perp \cap W^\perp), (U^\perp + W^\perp)$:

\[\begin{align*} \dim(U^\perp + W^\perp) &= \dim(U^\perp) + \dim(W^\perp) - \dim(U^\perp \cap W^\perp) \\\\ &= \dim(U^\perp) + \dim(W^\perp) - \dim((U \cap W)^\perp) \\\\ &= n - \dim(U) + n - \dim(W) - n + \dim(U \cap W) \\\\ &= n - (\dim(U) + \dim(W) - \dim(U \cap W)) \\\\ &= n - \dim(U \cup W) \\\\ &= \dim((U \cup W)^\perp) \end{align*}\]

(where $n = \dim V$).

5. $U \subseteq (U^\perp)^\perp$ with equality if $V$ is finite dimensional:

Let $u \in U$. Then $\forall w \in U^\perp$,

\[\langle u, w \rangle = \overline{\langle w, u\rangle} = 0\]

hence $\langle w, u \rangle = 0$ and $u \in (U^\perp)^\perp$. If $V$ is finite dimensional, by part 2:

\[\dim((U^\perp)^\perp) = \dim V - \dim U^\perp = \dim U\]

so $U = (U^\perp)^\perp$.



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