# The Sum of Cosine Squared in an Inner Product Space ## Problem 551

Let $\mathbf{v}$ be a vector in an inner product space $V$ over $\R$.
Suppose that $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is an orthonormal basis of $V$.
Let $\theta_i$ be the angle between $\mathbf{v}$ and $\mathbf{u}_i$ for $i=1,\dots, n$.

Prove that
$\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1.$ Add to solve later

## Definition (Angle between Vectors).

Let $\langle\mathbf{a}, \mathbf{b}\rangle$ denote the inner product of vectors $\mathbf{a}$ and $\mathbf{b}$ in $V$.

Recall that the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is defined as the unique number $\theta$ between $0$ and $\pi$ satisfying
$\cos \theta=\frac{\langle\mathbf{a}, \mathbf{b}\rangle}{\|\mathbf{a}\| \|\mathbf{b}\|}.$

## Proof.

Express the vector $\mathbf{v}$ as a linear combination of the basis vectors as
$\mathbf{v}=a_1\mathbf{u}_1+\dots+a_n\mathbf{u}_n$ for some real numbers $a_1, \dots, a_n$.

The length of the vector $\mathbf{v}$ is given by
$\|\mathbf{v}\|=\sqrt{a_1^2+\cdots+a_n^2}. \tag{*}$

For each $i$, we have using the properties of the inner product
\begin{align*}
\langle \mathbf{v}, \mathbf{u}_i\rangle&=\langle a_1\mathbf{u}_1+\dots+a_n\mathbf{u}_n, \mathbf{u}_i\rangle\\
&=a_1\langle\mathbf{u}_1, \mathbf{u}_i\rangle+\cdots +a_n \langle\mathbf{u}_n, \mathbf{u}_i \rangle\\
&=a_i \tag{**}
\end{align*}
since $\langle\mathbf{u}_i, \mathbf{u}_i\rangle=1$ and $\langle\mathbf{u}_j, \mathbf{u}_i\rangle=0$ if $j\neq i$ as $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is orthonormal.

By definition of the angle, we have
\begin{align*}
\cos \theta_i&=\frac{\langle\mathbf{v}, \mathbf{u}_i\rangle}{\|\mathbf{v}\| \|\mathbf{u}_i\|}=\frac{\langle\mathbf{v}, \mathbf{u}_i\rangle}{\|\mathbf{v}\| } && \text{since $\|\mathbf{u}_i\|=1$.}
\end{align*}
It follows that
\begin{align*}
\cos ^2\theta_1+\cdots+\cos^2 \theta_n &=\frac{\langle\mathbf{v}, \mathbf{u}_1\rangle^2}{\|\mathbf{v}\|^2 }+\cdots+\frac{\langle\mathbf{v}, \mathbf{u}_n\rangle^2}{\|\mathbf{v}\|^2 }\6pt] &=\frac{1}{\|\mathbf{v}\|^2}(a_1^2+\cdots a_n^2) &&\text{by (**)}\\[6pt] &=\frac{1}{\|\mathbf{v}\|^2}\cdot \|\mathbf{v}\|^2 &&\text{by (*)}\\[6pt] &=1. \end{align*} Thus we obtain \[\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1 as required. Add to solve later

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