
# 9.6 Orthogonal projections and minimization problems

Definition 9.6.1 Let $$V$$ be a finite-dimensional inner product space and $$U\subset V$$ be a subset (but not necessarily a subspace) of $$V$$. Then the orthogonal complement of $$U$$ is defined to be the set

$U^\bot = \{ v \in V \mid \inner{u}{v}=0 ~\rm{for~ all}~ u\in U \} .$

Note that, in fact, $$U^\bot$$ is always a subspace of $$V$$ (as you should check!) and that

$\{0\}^\bot = V \quad \text{and} \quad V^\bot = \{0\}.$

In addition, if $$U_{1}$$ and $$U_{2}$$ are subsets of $$V$$ satisfying $$U_1\subset U_2$$, then $$U_2^\bot \subset U_1^\bot$$.     Remarkably, if $$U\subset V$$ is a subspace of $$V$$, then we can say quite a bit more about $$U^{\bot}$$.

Theorem 9.6.2. If $$U\subset V$$ is a subspace of $$V$$, then $$V=U\oplus U^\bot$$.

Proof. We need to show two things:

1. $$V=U+U^\bot$$.
2. $$U\cap U^\bot = \{0\}$$.

To show Condition~1 holds, let $$(e_1,\ldots,e_m)$$ be an orthonormal basis of $$U$$. Then, for all $$v\in V$$, we can write
\label{eq:U Ubot}
v=\underbrace{\inner{v}{e_1} e_1 + \cdots + \inner{v}{e_m} e_m}_u +
\underbrace{v-\inner{v}{e_1} e_1 - \cdots - \inner{v}{e_m} e_m}_w.  \tag{9.6.1}

The vector $$u\in U$$, and
\begin{equation*}
\inner{w}{e_j} = \inner{v}{e_j} - \inner{v}{e_j} =0, ~\rm{for~ all}~  {j=1,2,\ldots,m  , }
\end{equation*}
since $$(e_1,\ldots,e_m)$$ is an orthonormal list of vectors. Hence, $$w\in U^\bot$$. This implies that $$V=U+U^\bot$$.

To prove that Condition~2 also holds, let $$v\in U\cap U^\bot$$. Then $$v$$ has to be orthogonal to every vector in $$U$$, including to itself, and so $$\inner{v}{v}=0$$. However, this implies $$v=0$$ so that $$U\cap U^\bot=\{0\}$$.

Example 9.6.3.  $$\mathbb{R}^2$$ is the direct sum of any two orthogonal lines, and $$\mathbb{R}^3$$ is the direct sum of any plane and any line orthogonal to the plane as illustrated in Figure 9.6.1. For example,
\begin{equation*}
\begin{split}
\mathbb{R}^2 &= \{(x,0) \mid x\in \mathbb{R}\} \oplus \{ (0,y) \mid y\in\mathbb{R}\},\\
\mathbb{R}^3 &= \{(x,y,0) \mid x,y\in \mathbb{R}\} \oplus \{(0,0,z) \mid z\in \mathbb{R}\}.
\end{split}
\end{equation*}

Figure 9.6.1: $$\mathbb{R^3}$$ as a direct sum of a plane and a line.

Another fundamental fact about the orthogonal complement of a subspace is as follows.

Theorem 9.6.4. If $$U\subset V$$ is a subspace of $$V$$, then $$U=(U^\bot)^\bot$$.

Proof.  First we show that $$U\subset (U^\bot)^\bot$$. Let $$u\in U$$. Then, for all $$v\in U^\bot$$, we have $$\inner{u}{v}=0$$. Hence, $$u\in (U^\bot)^\bot$$ by the definition of $$(U^\bot)^\bot$$.

Next we show that $$(U^\bot)^\bot\subset U$$. Suppose $$0\neq v\in (U^\bot)^\bot$$ such that $$v\not\in U$$, and decompose $$v$$ according to Theorem 9.6.2, i.e., as
\begin{equation*}
v = u_1 + u_2 \in U \oplus U^\bot
\end{equation*}
with $$u_1\in U$$ and $$u_2\in U^\bot$$. Then $$u_2\neq 0$$ since $$v\not\in U$$.

Furthermore, $$\inner{u_2}{v} = \inner{u_2}{u_2} \neq 0$$. But then $$v$$ is not in $$(U^\bot)^\bot$$, which contradicts our initial assumption. Hence, we must have that $$(U^\bot)^\bot\subset U$$.

By Theorem 9.6.2, we have the decomposition $$V=U\oplus U^\bot$$ for every subspace $$U\subset V$$. This allows us to define the orthogonal projection $$P_U$$ of $$V$$ onto $$U$$.

Definition 9.6.5.  Let $$U\subset V$$ be a subspace of a finite-dimensional inner product space. Every $$v\in V$$ can be uniquely written as $$v=u+w$$ where $$u\in U$$ and $$w\in U^\bot$$. Define
\begin{equation*}
\begin{split}
P_U:V &\to V,\\
v &\mapsto u.
\end{split}
\end{equation*}

Note that $$P_U$$ is called a projection operator since it satisfies $$P_U^2=P_U$$. Further, since we also have
\begin{equation*}
\begin{split}
&\range(P_U)= U,\\
&\kernel(P_U) = U^\bot,
\end{split}
\end{equation*}
it follows that $$\range(P_U) \bot \kernel(P_U)$$. Therefore, $$P_U$$ is called an orthogonal projection.

The decomposition of a vector $$v\in V$$ as given in Equation (9.6.1) yields the formula
\label{eq:ortho decomp}
P_U v = \inner{v}{e_1} e_1 + \cdots + \inner{v}{e_m} e_m,  \tag{9.6.2}

where $$(e_1,\ldots,e_m)$$ is any orthonormal basis of $$U$$. Equation (9.6.2) is a particularly useful tool for computing such things as the matrix of $$P_{U}$$ with respect to the basis $$(e_1,\ldots,e_m)$$.

Let us now apply the inner product to the following minimization problem: Given a subspace $$U\subset V$$ and a vector $$v\in V$$, find the vector $$u\in U$$ that is closest to the vector $$v$$. In other words, we want to make $$\norm{v-u}$$ as small as possible. The next proposition shows that $$P_Uv$$ is the closest point in $$U$$ to the vector $$v$$ and that this minimum is, in fact, unique.

Proposition 9.6.6.  Let $$U\subset V$$ be a subspace of $$V$$ and $$v\in V$$. Then

$\norm{v-P_Uv} \le \norm{v-u} \qquad \text{for every $$u\in U$$.}$

Furthermore, equality holds if and only if $$u=P_Uv$$.

Proof.  Let $$u\in U$$ and set $$P:=P_U$$ for short. Then
\begin{equation*}
\begin{split}
\norm{v-P v}^2 &\le \norm{v-P v}^2 + \norm{Pv-u}^2\\
&= \norm{(v-P v)+(P v-u)}^2 = \norm{v-u}^2,
\end{split}
\end{equation*}
where the second line follows from the Pythagorean Theorem 9.3.2~\ref{thm:pythagoras} since $$v-Pv\in U^\bot$$ and $$Pv-u\in U$$. Furthermore, equality holds only if $$\norm{Pv-u}^2=0$$, which is equivalent to $$Pv=u$$.

Example 9.6.7. Consider the plane $$U\subset \mathbb{R}^3$$ through 0 and perpendicular to the vector $$u=(1,1,1)$$. Using the standard norm on $$\mathbb{R}^3$$, we can calculate the distance of the point $$v=(1,2,3)$$ to $$U$$ using Proposition 9.6.6. In particular, the distance $$d$$ between $$v$$ and $$U$$ is given by $$d=\norm{v-P_Uv}$$. Let $$(\frac{1}{\sqrt{3}}u,u_1,u_2)$$ be a basis for $$\mathbb{R}^3$$ such that $$(u_1,u_2)$$ is an orthonormal basis of $$U$$. Then, by Equation (9.6.2), we have
\begin{align*}
v-P_Uv & = (\frac{1}{3}\inner{v}{u}u+\inner{v}{u_1}u_1+\inner{v}{u_2}u_2) - (\inner{v}{u_1}u_1 +\inner{v}{u_2}u_2)\\
& = \frac{1}{3}\inner{v}{u}u\\
& = \frac{1}{3}\inner{(1,2,3)}{(1,1,1)}(1,1,1)\\
& = (2,2,2).
\end{align*}
Hence, $$d=\norm{(2,2,2)}=2\sqrt{3}$$.

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