<Home>Teaching><MATH 115A (Spring 2023)>Worksheet 9

Worksheet 9

Problem 3

Let $V$ be an inner product space, let $W$ be a finite-dimensional subspace of $V$ . Prove that if $x \notin W$ , then there exists $y \in W^\perp$ with $\inner{x, y} \neq 0$ .

Solution.

Recall that $V = W \oplus W^\perp$ (via orthogonal projection), so we can write

x = x_W + x_{W^\perp},

where $x_W \in W$ and $x_{W^\perp} \in W^\perp$ . Note that $x_{W^\perp} \neq 0$ : if this were not the case, then $x = x_W \in W$ , which is impossible. Thus, if we set $y = x_{W^\perp}$ , then

\begin{aligned} \inner{x, y} &= \inner{x_W + y, y} \\ &= \inner{x_W, y} + \inner{y, y} \\ &= \norm{y}^2 \\ &> 0. \end{aligned}

The penultimate inequality follows from the fact that $y$ is orthogonal to every vector in $W$ , and $\norm{y}$ is strictly larger than $0$ since $y \neq 0$ .

Problem 4

Let $V$ be a finite-dimensional inner product space, let $W$ be a subspace of $V$ . Prove that $V/W$ is isomorphic to $W^\perp$ .

Solution.

There are two different ways to do this:

Every finite-dimensional vector space over a field $\F$ is isomorphic to $\F^n$ for some $n$ , so we can try proving $\dim\p{V/W} = \dim\p{W^\perp}$ .
We can try to find a linear isomorphism $T\colon W^\perp \to V/W$ .

Method 1

Let $\dim V = n$ and $\dim W = m$ . Recall that because $V, W$ are both finite-dimensional that $\dim\p{V/W} = n - m$ . To show $\dim\p{W^\perp} = n - m$ , we need to find a basis for $W^\perp$ .

Let $\set{v_1, \ldots, v_m}$ be an orthonormal basis (it's very important that we pick an orthonormal basis for this to work) for $W$ , and extend it into an orthonormal basis $\beta = \set{v_1, \ldots, v_m, \ldots, v_n}$ for $V$ . We will show that $\set{v_{m+1}, \ldots, v_n}$ is a basis for $W^\perp$ .

First, the set $\set{v_{m+1}, \ldots, v_n}$ is linearly independent since $\beta$ was. Next, we need to show that it spans $W^\perp$ . Let $v \in W^\perp$ . Since $\beta$ was an orthonormal basis for $V$ , we can write

v = \inner{v, v_1}v_1 + \cdots + \inner{v, v_m}v_m + \cdots + \inner{v, v_n}v_n.

But $v \in W^\perp$ and $v_1, \ldots, v_m \in W^\perp$ , so $\inner{v, v_1} = \cdots = \inner{v, v_m} = 0$ . Thus,

v = \inner{v, v_{m+1}}v_{m+1} + \cdots + \inner{v, v_n}v_n.

This means $v \in \span\set{v_{m+1}, \ldots, v_n}$ , so the set spans. Thus, this set is a basis for $W^\perp$ and has $n - m$ vectors, so $\dim W^\perp = n - m$ .

Method 2

Define $T\colon W^\perp \to V/W$ by

T\p{v} = v + W.

This map is linear: given $u, v \in W^\perp$ and a scalar $c \in \F$ , the definitions of addition and scalar multiplication in $V/W$ tell us

\begin{aligned} T\p{cu + v} &= \p{cu + v} + W \\ &= c\p{u + W} + \p{v + W} \\ &= cT\p{u} + T\p{v}, \end{aligned}

so $T$ is linear. Let's show that $T$ is one-to-one: let $v \in N\p{T}$ . Then

v + W = T\p{v} = W,

so $v \in W$ . But $v \in W^\perp$ , so $v \in W \cap W^\perp = \set{\vec{0}}$ , and so $v = \vec{0}$ . This shows that $N\p{T} = \set{0}$ , so $T$ is one-to-one.

Now let's show that $T$ is onto. Let $v + W \in V/W$ . Note that we can't just say that

T\p{v} = v + W.

This is because $v$ does not have to be in $W^\perp$ . However, if we could replace $v$ with a vector $y \in W^\perp$ such that $v + W = y + W$ , then $T\p{y} = v + W$ and we are done. The "natural" way to do this is to write

v = v_W + v_{W^\perp}

where $v_W \in W$ and $v_{W^\perp} \in W^\perp$ as in Problem 3. Thus, by definition of $v + W$ , we have

v + W = \p{v_W + v_{W^\perp}} + W = v_{W^\perp} + W,

since $v_W \in W$ . Hence, we have

T\p{v_{W^\perp}} = v_{W^\perp} + W = v + W,

so $T$ is onto.

We have shown that $T$ is an invertible linear map, so it's automatically an isomorphism.