# 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 $V$.

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## 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 $V$.

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