# Normalize Lengths to Obtain an Orthonormal Basis

## Problem 715

Let
$\mathbf{v}_{1} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} ,\; \mathbf{v}_{2} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} .$ Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?

If not, then find an orthonormal basis for $V$.

## Solution.

We begin by computing
\begin{align*}
\mathbf{v}_{1}\cdot\mathbf{v}_{2}
&=
\mathbf{v}_{1}^{T}\mathbf{v}_{2}
=
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
=
1\cdot 1+1\cdot -1
=
1-1
=
0,
\\
\mathbf{v}_{1}\cdot\mathbf{v}_{1}
&=
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
=
1+1
=
2,
\\
\mathbf{v}_{2}\cdot\mathbf{v}_{2}
&=
\begin{bmatrix}
1 & -1
\end{bmatrix}
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
=
1+1
=
2.
\end{align*}
Since $\mathbf{v}_{1}\cdot\mathbf{v}_{2}=0$, the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are orthogonal. Since $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are nonzero orthogonal vectors, they are linearly independent, and it follows that $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthogonal basis for $V$. However, since $\mathbf{v}_{i}\cdot\mathbf{v}_{i}=2\neq 1$ for $i=1,2$, we know that $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ do not form an orthonormal basis for $V$.

To find an orthonormal basis for $V$, note that for any scalars $a$ and $b$, $(a\mathbf{v}_{1})\cdot(b\mathbf{v}_{2})=ab(\mathbf{v}_{1}\cdot\mathbf{v}_{2})=ab\cdot 0=0$. Therefore, $a\mathbf{v}_{1}$ and $b\mathbf{v}_{2}$ will always form an orthogonal basis for $V$. All we need to do is choose $a$ and $b$ so that $a\mathbf{v}_{1}$ and $b\mathbf{v}_{2}$ form an orthonormal set. For $a$, we require
$1 =\|a\mathbf{v}\|=a\|\mathbf{v}\|$ and so
$a=\frac{1}{\|\mathbf{v}\|}=\frac{1}{\sqrt{2}}.$ (Note that $\|\mathbf{v}\| \neq 0$ as $\mathbf{v}\neq \mathbf{0}$. Also, note that to obtain a length 1 vector, we just needed divide the vector by its length.)
Similarly, $b=1/\sqrt{2}$. Therefore, if we define
\begin{align*}
\mathbf{w}_{1}
&=
\dfrac{1}{\sqrt{2}}
\mathbf{v}_{1}
=
\dfrac{1}{\sqrt{2}}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
,
\\
\mathbf{w}_{2}
&=
\dfrac{1}{\sqrt{2}}
\mathbf{v}_{2}
=
\dfrac{1}{\sqrt{2}}
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
,
\end{align*}
then $\mathbf{w}_{1}$ and $\mathbf{w}_{2}$ form an orthonormal basis for $V$.

Let $W$ be the set of $3\times 3$ skew-symmetric matrices. Show that $W$ is a subspace of the vector space...