# 9.3 Orthogonality

- Page ID
- 258

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

Using the inner product, we can now define the notion of orthogonality, prove that the Pythagorean theorem holds in any inner product space, and use the Cauchy-Schwarz inequality to prove the triangle inequality. In particular, this will show that \(\norm{v}=\sqrt{\inner{v}{v}}\) does indeed define a norm.

**Definition 9.3.1. **Two vectors \(u,v\in V \) are **orthogonal** (denoted \(u\bot v\)) if \(\inner{u}{v}=0\).

Note that the zero vector is the only vector that is orthogonal to itself. In fact, the zero vector is orthogonal to every vector \(v\in V\).

**Theorem 9.3.2.** (Pythagorean Theorem). *If *\(u,v\in V\), *an inner product space, with* \(u\bot v\), *then* \(\norm{\cdot} \) *defined by* \(\norm{v}:=\sqrt{\inner{v}{v}} \) *obeys*

\[ \norm{u+v}^2 = \norm{u}^2 + \norm{v}^2. \]

*Proof.* Suppose \(u,v\in V \) such that \(u\bot v\). Then

\begin{equation*}

\begin{split}

\norm{u+v}^2 &= \inner{u+v}{u+v}

= \norm{u}^2 + \norm{v}^2 + \inner{u}{v} + \inner{v}{u}\\

&= \norm{u}^2 + \norm{v}^2.

\end{split}

\end{equation*}

Note that the converse of the Pythagorean Theorem holds for real vector spaces since, in that case, \(\inner{u}{v}+\inner{v}{u}=2\mathrm{Re} \inner{u}{v} =0\).

Given two vectors \(u,v\in V \) with \(v\neq 0\), we can uniquely decompose \(u \) into two pieces: one piece parallel to \(v \) and one piece orthogonal to \(v\). This is called an **orthogonal decomposition**. More precisely, we have

\begin{equation*}

u=u_1+u_2,

\end{equation*}

where \(u_1=a v \) and \(u_2\bot v \) for some scalar \(a \in \mathbb{F}\). To obtain such a decomposition, write \(u_2=u-u_1=u-av\). Then, for \(u_2 \) to be

orthogonal to \(v\), we need

\begin{equation*}

0 = \inner{u-av}{v} = \inner{u}{v} - a\norm{v}^2.

\end{equation*}

Solving for \(a \) yields \(a=\inner{u}{v}/\norm{v}^2 \) so that

\begin{equation}\label{eq:orthogonal decomp}

u=\frac{\inner{u}{v}}{\norm{v}^2} v + \left( u-\frac{\inner{u}{v}}{\norm{v}^2} v \right). \tag{9.3.1}

\end{equation}

This decomposition is particularly useful since it allows us to provide a simple proof for the Cauchy-Schwarz inequality.

**Theorem 9.3.3 **(Cauchy-Schwarz inequality). *Given any* \(u,v\in V\), *we have*

\[ |\inner{u}{v}| \le \norm{u} \norm{v}. \]

*Furthermore, equality holds if and only if* \(u \) *and* \(v \) *are linearly dependent, i.e., are scalar multiples of each other.*

*Proof. *If \(v=0\), then both sides of the inequality are zero. Hence, assume that \(v\neq 0\), and consider the orthogonal decomposition

\begin{equation*}

u = \frac{\inner{u}{v}}{\norm{v}^2} v + w

\end{equation*}

where \(w\bot v\). By the Pythagorean theorem, we have

\begin{equation*}

\norm{u}^2 = \left\| \frac{\inner{u}{v}}{\norm{v}^2}v \right\|^2 + \norm{w}^2

= \frac{|\inner{u}{v}|^2}{\norm{v}^2} + \norm{w}^2 \ge \frac{|\inner{u}{v}|^2}{\norm{v}^2}.

\end{equation*}

Multiplying both sides by \(\norm{v}^2 \) and taking the square root then yields the Cauchy-Schwarz inequality.

Note that we get equality in the above arguments if and only if \(w=0\). But, by Equation (9.3.1), this means that \(u \) and \(v \) are linearly dependent.

The Cauchy-Schwarz inequality has many different proofs. Here is another one.

*Alternate Proof of Theorem 9.3.3. *Given \(u,v\in V\), consider the norm square of the vector \(u+r e^{i\theta}v:\)

\begin{equation*}

0 \le \norm{u+re^{i\theta}v}^2

= \norm{u}^2 + r^2 \norm{v}^2 + 2 \mathrm{Re} (r e^{i\theta} \inner{u}{v}).

\end{equation*}

Since \( \inner{u}{v} \) is a complex number, one can choose \(\theta \) so that \(e^{i \theta} \inner{u}{v} \) is real. Hence, the right hand side is a parabola \(a r^2 +b r +c \) with real coefficients. It will lie above the real axis, i.e. \(ar^2+br+c\ge 0\), if it does not have any real solutions for \(r\). This is the case when the discriminant satisfies \(b^2-4ac\le 0\). In our case this means

\begin{equation*}

4 | \inner{u}{v} |^2 -4 \norm{u}^2 \norm{v}^2\le 0.

\end{equation*}

Moreover, equality only holds if \(r \) can be chosen such that \(u+r e^{i\theta} v =0\), which means that \(u \) and \(v \) are scalar multiples.

Now that we have proven the Cauchy-Schwarz inequality, we are finally able to verify the triangle inequality. This is the final step in showing that \(\norm{v}=\sqrt{\inner{v}{v}}\) does indeed define a norm. We illustrate the triangle inequality in Figure 9.3.1.

**Figure 9.3.1**: *The Triangle Inequality in* \( \mathbb{R^2} \)

**Theorem 9.3.4.** (Triangle Inequality). *For all* \(u,v\in V \) *we have*

\[ \norm{u+v} \le \norm{u} + \norm{v}. \]

*Proof. *By a straightforward calculation, we obtain

\begin{equation*}

\begin{split}

\norm{u+v}^2 &= \inner{u+v}{u+v} = \inner{u}{u} + \inner{v}{v} +\inner{u}{v} + \inner{v}{u}\\

&= \inner{u}{u} + \inner{v}{v} + \inner{u}{v} +\overline{\inner{u}{v}}

= \norm{u}^2 + \norm{v}^2 +2 \mathrm{Re} \inner{u}{v}.

\end{split}

\end{equation*}

Note that \(\mathrm{Re}\inner{u}{v}\le |\inner{u}{v}| \) so that, using the Cauchy-Schwarz inequality, we obtain

\begin{equation*}

\norm{u+v}^2 \le \norm{u}^2 + \norm{v}^2 + 2\norm{u}\norm{v} = (\norm{u}+\norm{v})^2.

\end{equation*}

Taking the square root of both sides now gives the triangle inequality.

*Remark* 9.3.5. Note that equality holds for the triangle inequality if and only if \(v=ru \) or \(u=rv \) for some \(r\ge 0\). Namely, equality in the proof happens only if \(\inner{u}{v} =\norm{u}\norm{v}\), which is equivalent to \(u \) and \(v \) being scalar multiples of one another.

**Theorem 9.3.6.** (Parallelogram Law). *Given any* \(u,v\in V\), *we have*

\[ \norm{u+v}^2 + \norm{u-v}^2 = 2(\norm{u}^2 + \norm{v}^2). \]

*Proof. *By direct calculation,

\begin{equation*}

\begin{split}

\norm{u+v}^2 + \norm{u-v}^2 &= \inner{u+v}{u+v} + \inner{u-v}{u-v}\\

&= \norm{u}^2 + \norm{v}^2 + \inner{u}{v} + \inner{v}{u} + \norm{u}^2 + \norm{v}^2

- \inner{u}{v} -\inner{v}{u}\\

&=2(\norm{u}^2 + \norm{v}^2).

\end{split}

\end{equation*}

*Remark 9.3.7. *We illustrate the parallelogram law in Figure 9.3.2.

**Figure 9.3.2:** *The Parallelogram Law in* \( \mathbb{R^2} \)

### Contributors

- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.