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6.3 Gram-Schmidt Process: Class Notes

Definition

A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthonormal.

Just as with the Euclidean inner product, we can normalize a nonzero vector $𝐯$ to obtain a unit vector:

$${\displaystyle 𝐮=\frac{1}{\Vert 𝐯\Vert }𝐯}$$

Theorem 6.3.1

Proof

Suppose $k_1{𝐯}_1+k_2{𝐯}_2+\cdots +k_n{𝐯}_n=\mathrm{𝟎}$ and consider, for $1\le i\le n$,

$$⟨k_1{𝐯}_1+k_2{𝐯}_2+\cdots +k_n{𝐯}_n,{𝐯}_i⟩=⟨\mathrm{𝟎},{𝐯}_i⟩=0$$

But by the additivity and homogeneity axioms,

$$\begin{array}{cc}\hfill {\displaystyle ⟨k_1{𝐯}_1+k_2{𝐯}_2+\cdots +k_n{𝐯}_n,{𝐯}_i⟩}& {\displaystyle =⟨k_1{𝐯}_1,{𝐯}_i⟩+⟨k_2{𝐯}_2,{𝐯}_i⟩+\cdots +⟨k_i{𝐯}_i,{𝐯}_i⟩+\cdots +⟨k_n{𝐯}_n,{𝐯}_i⟩}\hfill \\ \hfill {\displaystyle }& {\displaystyle =⟨k_i{𝐯}_i,{𝐯}_i⟩=k_i⟨{𝐯}_i,{𝐯}_i⟩}\hfill \end{array}$$

since $⟨{𝐯}_k,{𝐯}_i⟩=0$ for $k\ne i$ because $S$ is orthogonal. So $k_i⟨{𝐯}_i,{𝐯}_i⟩=0$.

Since ${𝐯}_i\ne \mathrm{𝟎}$, $⟨{𝐯}_i,{𝐯}_i⟩\ne 0$ by the positivity axiom, so $k_i=0$ for every $i$. Thus $S$ is linearly independent.

Definition

In an inner product space, a basis comprising orthogonal vectors is an orthogonal basis, and a basis comprising orthonormal vectors is an orthonormal basis.

Example: With the Euclidean inner product, $\{(3,4,0),(-4,3,0),(0,0,1)\}$ is an orthogonal basis for $R^3$. Normalizing each vector gives an orthonormal basis:

$${\displaystyle \{(\frac{3}{5},\frac{4}{5},0),(-\frac{4}{5},\frac{3}{5},0),(0,0,1)\}}$$

#6

Show that the column vectors of $A$ form an orthogonal basis for the column space of $A$ with respect to the Euclidean inner product, and then find an orthonormal basis for that column space.

$$A=\left[\begin{array}{ccc}1/5& -1/2& 1/3\\ 1/5& 1/2& 1/3\\ 1/5& 0& -2/3\end{array}\right]$$

Solution

Call the columns ${\overrightarrow{c}}_1$, ${\overrightarrow{c}}_2$, and ${\overrightarrow{c}}_3$. Check each pair:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {\overrightarrow{c}}_1·{\overrightarrow{c}}_2}& {\displaystyle =(\frac{1}{5},\frac{1}{5},\frac{1}{5})·(-\frac{1}{2},\frac{1}{2},0)=-\frac{1}{10}+\frac{1}{10}+0=0}\hfill \\ \hfill {\displaystyle {\overrightarrow{c}}_1·{\overrightarrow{c}}_3}& {\displaystyle =(\frac{1}{5},\frac{1}{5},\frac{1}{5})·(\frac{1}{3},\frac{1}{3},-\frac{2}{3})=\frac{1}{15}+\frac{1}{15}-\frac{2}{15}=0}\hfill \\ \hfill {\displaystyle {\overrightarrow{c}}_2·{\overrightarrow{c}}_3}& {\displaystyle =(-\frac{1}{2},\frac{1}{2},0)·(\frac{1}{3},\frac{1}{3},-\frac{2}{3})=-\frac{1}{6}+\frac{1}{6}+0=0}\hfill \end{array}}$$

The columns are nonzero and orthogonal, so by Theorem 6.3.1 they are linearly independent. They span the column space, so they form an orthogonal basis for it.

To normalize, first scale the columns to clear the fractions (scaling by a positive number doesn’t change the unit vector):

$$\begin{array}{cc}\hfill {\displaystyle 5{\overrightarrow{c}}_1}& {\displaystyle =(1,1,1),\Vert 5{\overrightarrow{c}}_1\Vert =\sqrt{3}}\hfill \\ \hfill {\displaystyle 2{\overrightarrow{c}}_2}& {\displaystyle =(-1,1,0),\Vert 2{\overrightarrow{c}}_2\Vert =\sqrt{2}}\hfill \\ \hfill {\displaystyle 3{\overrightarrow{c}}_3}& {\displaystyle =(1,1,-2),\Vert 3{\overrightarrow{c}}_3\Vert =\sqrt{6}}\hfill \end{array}$$

Divide each scaled column by its norm. The orthonormal basis is

Theorem 6.3.2

So with an orthogonal basis, the coordinates of $𝐮$ come straight from inner products, with no system to solve: the boxed coefficients in (a) are the entries of the coordinate vector.

$${\displaystyle 𝐮=\frac{⟨𝐮,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1+\frac{⟨𝐮,{𝐯}_2⟩}{\Vert {𝐯}_2{\Vert }^2}{𝐯}_2+\cdots +\frac{⟨𝐮,{𝐯}_n⟩}{\Vert {𝐯}_n{\Vert }^2}{𝐯}_n}$$

$${\displaystyle (𝐮{)}_S=(\frac{⟨𝐮,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2},\frac{⟨𝐮,{𝐯}_2⟩}{\Vert {𝐯}_2{\Vert }^2},\dots ,\frac{⟨𝐮,{𝐯}_n⟩}{\Vert {𝐯}_n{\Vert }^2})}$$

Proof

Let $𝐮$ be in $V$. Since $S$ is a basis for $V$, there are scalars $c_i$ with $𝐮=c_1{𝐯}_1+c_2{𝐯}_2+\cdots +c_n{𝐯}_n$. For $1\le i\le n$,

$$⟨𝐮,{𝐯}_i⟩=⟨c_1{𝐯}_1+c_2{𝐯}_2+\cdots +c_n{𝐯}_n,{𝐯}_i⟩=c_i⟨{𝐯}_i,{𝐯}_i⟩=c_i\Vert {𝐯}_i{\Vert }^2$$

because $S$ is orthogonal (every other term is 0, as in the proof of Theorem 6.3.1). Hence

$${\displaystyle c_i=\frac{⟨𝐮,{𝐯}_i⟩}{\Vert {𝐯}_i{\Vert }^2}}$$

If $S$ is an orthonormal basis, then $\Vert {𝐯}_i{\Vert }^2=1$, so $c_i=⟨𝐮,{𝐯}_i⟩$. This is part (b).

#8

Use Theorem 6.3.2(b) to express the vector $𝐮=(3,-7,4)$ as a linear combination of the vectors ${\displaystyle {𝐯}_1=(-\frac{3}{5},\frac{4}{5},0)}$, ${\displaystyle {𝐯}_2=(\frac{4}{5},\frac{3}{5},0)}$, ${𝐯}_3=(0,0,1)$.

Solution

First note that this is an orthonormal basis with respect to the Euclidean inner product (the dot product): ${\displaystyle {𝐯}_1·{𝐯}_2=-\frac{12}{25}+\frac{12}{25}=0}$, ${𝐯}_1·{𝐯}_3={𝐯}_2·{𝐯}_3=0$, and each vector has norm 1. So

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 𝐮}& {\displaystyle =⟨𝐮,{𝐯}_1⟩{𝐯}_1+⟨𝐮,{𝐯}_2⟩{𝐯}_2+⟨𝐮,{𝐯}_3⟩{𝐯}_3}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-\frac{9}{5}-\frac{28}{5}){𝐯}_1+(\frac{12}{5}-\frac{21}{5}){𝐯}_2+4{𝐯}_3}\hfill \end{array}}$$

#12

Find the coordinate vector $(𝐮{)}_S$ for the vector $𝐮$ and the basis $S$ that were given in Exercise 8.

Solution

The coordinates are the coefficients found in #8.

Note that since $S$ is an orthonormal basis, no augmented matrix is required.

Theorem 6.3.3 Projection Theorem (generalization of Theorem 3.3.2)

Definition

If $W$ is a finite-dimensional subspace of an inner product space $V$ and a vector $𝐮$ in $V$ is expressed as $𝐮={𝐰}_1+{𝐰}_2$, where ${𝐰}_1$ is in $W$ and ${𝐰}_2$ is in $W^{⟂}$, then ${𝐰}_1$ is called the orthogonal projection of $𝐮$ on $𝐖$ and is denoted by ${𝐰}_1={\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_W𝐮$ and ${𝐰}_2$ is called the orthogonal projection of $𝐮$ on ${𝐖}^{⟂}$ and is denoted by ${𝐰}_2={\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W^{⟂}}𝐮$. The vector ${𝐰}_2$ is also called the component of $𝐮$ orthogonal to $𝐖$.

Theorem 6.3.4

Here $V$ is $n$-dimensional and $W$ is $r$-dimensional. The basis vectors ${𝐯}_1,\dots ,{𝐯}_r$ lie in $W$ (a basis for $V$ would continue on to ${𝐯}_n$), and $𝐮$ is any vector in $V$, not necessarily in $W$.

Proof

From Theorem 6.3.3, $𝐮={𝐰}_1+{𝐰}_2$ with ${𝐰}_1$ in $W$ and ${𝐰}_2$ in $W^{⟂}$. Since ${𝐰}_1$ is in $W$, Theorem 6.3.2(a) applied to $W$ gives

$${\displaystyle {\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_W𝐮={𝐰}_1=\frac{⟨{𝐰}_1,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1+\frac{⟨{𝐰}_1,{𝐯}_2⟩}{\Vert {𝐯}_2{\Vert }^2}{𝐯}_2+\cdots +\frac{⟨{𝐰}_1,{𝐯}_r⟩}{\Vert {𝐯}_r{\Vert }^2}{𝐯}_r}$$

Since ${𝐰}_2$ is in $W^{⟂}$, $⟨{𝐰}_2,{𝐯}_i⟩=0$, so

$$⟨{𝐰}_1,{𝐯}_i⟩=⟨{𝐰}_1,{𝐯}_i⟩+⟨{𝐰}_2,{𝐯}_i⟩=⟨{𝐰}_1+{𝐰}_2,{𝐯}_i⟩=⟨𝐮,{𝐯}_i⟩$$

Therefore

$${\displaystyle {\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_W𝐮=\frac{⟨𝐮,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1+\cdots +\frac{⟨𝐮,{𝐯}_r⟩}{\Vert {𝐯}_r{\Vert }^2}{𝐯}_r=\sum _{i=1}^r\frac{⟨𝐮,{𝐯}_i⟩}{\Vert {𝐯}_i{\Vert }^2}{𝐯}_i}$$

If the basis is orthonormal, then $\Vert {𝐯}_i{\Vert }^2=1$, which gives part (b).

#24

The vectors ${𝐯}_1=(0,1,-4,-1)$ and ${𝐯}_2=(3,5,1,1)$ are orthogonal with respect to the Euclidean inner product on $R^4$. Find the orthogonal projection ${𝐰}_1$ of $𝐛=(1,2,0,-2)$ on the subspace $W$ spanned by these vectors. [Extension of the exercise] Then find ${𝐰}_2$ in $W^{⟂}$ such that $𝐛={𝐰}_1+{𝐰}_2$.

Solution

By Theorem 6.3.4(a), with $𝐛·{𝐯}_1=0+2+0+2=4$, $\Vert {𝐯}_1{\Vert }^2=0+1+16+1=18$, $𝐛·{𝐯}_2=3+10+0-2=11$, and $\Vert {𝐯}_2{\Vert }^2=9+25+1+1=36$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐰}_1={\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_W𝐛}& {\displaystyle =\frac{𝐛·{𝐯}_1}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1+\frac{𝐛·{𝐯}_2}{\Vert {𝐯}_2{\Vert }^2}{𝐯}_2}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{4}{18}(0,1,-4,-1)+\frac{11}{36}(3,5,1,1)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{8}{36}(0,1,-4,-1)+\frac{11}{36}(3,5,1,1)=\frac{1}{36}(33,63,-21,3)}\hfill \end{array}}$$

$${\displaystyle {𝐰}_2=𝐛-{𝐰}_1=(1,2,0,-2)-(\frac{11}{12},\frac{7}{4},-\frac{7}{12},\frac{1}{12})}$$

Check that ${𝐰}_2$ is in $W^{⟂}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐰}_2·{𝐯}_1}& {\displaystyle =0+\frac{3}{12}-\frac{28}{12}+\frac{25}{12}=0}\hfill \\ \hfill {\displaystyle {𝐰}_2·{𝐯}_2}& {\displaystyle =\frac{3}{12}+\frac{15}{12}+\frac{7}{12}-\frac{25}{12}=0}\hfill \end{array}}$$

#25

The vectors ${𝐯}_1$, ${𝐯}_2$, and ${𝐯}_3$ are orthonormal with respect to the Euclidean inner product on $R^4$. Find the orthogonal projection of $𝐛=(1,2,0,-1)$ onto the subspace $W$ spanned by these vectors.

${\displaystyle {𝐯}_1=(0,\frac{1}{\sqrt{18}},-\frac{4}{\sqrt{18}},-\frac{1}{\sqrt{18}})}$, ${\displaystyle {𝐯}_2=(\frac{1}{2},\frac{5}{6},\frac{1}{6},\frac{1}{6})}$, ${\displaystyle {𝐯}_3=(\frac{1}{\sqrt{18}},0,\frac{1}{\sqrt{18}},-\frac{4}{\sqrt{18}})}$

Solution

The basis is orthonormal, so by Theorem 6.3.4(b),

$${\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_W𝐛=⟨𝐛,{𝐯}_1⟩{𝐯}_1+⟨𝐛,{𝐯}_2⟩{𝐯}_2+⟨𝐛,{𝐯}_3⟩{𝐯}_3$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ⟨𝐛,{𝐯}_1⟩}& {\displaystyle =\frac{0+2+0+1}{\sqrt{18}}=\frac{3}{\sqrt{18}}}\hfill \\ \hfill {\displaystyle ⟨𝐛,{𝐯}_2⟩}& {\displaystyle =\frac{1}{2}+\frac{10}{6}+0-\frac{1}{6}=\frac{12}{6}=2}\hfill \\ \hfill {\displaystyle ⟨𝐛,{𝐯}_3⟩}& {\displaystyle =\frac{1+0+0+4}{\sqrt{18}}=\frac{5}{\sqrt{18}}}\hfill \end{array}}$$

Multiply out (for example, ${\displaystyle \frac{3}{\sqrt{18}}{𝐯}_1=\frac{3}{18}(0,1,-4,-1)}$):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_W𝐛}& {\displaystyle =(0,\frac{1}{6},-\frac{2}{3},-\frac{1}{6})+(1,\frac{5}{3},\frac{1}{3},\frac{1}{3})}\hfill \\ \hfill {\displaystyle }& {\displaystyle +(\frac{5}{18},0,\frac{5}{18},-\frac{10}{9})}\hfill \end{array}}$$

Theorem 6.3.5 (proof outlines the Gram-Schmidt process)

Proof

This is a constructive proof. Let $W$ be any nonzero finite-dimensional subspace of an inner product space and suppose $\{{𝐮}_1,{𝐮}_2,\dots ,{𝐮}_r\}$ is any basis for $W$. We will produce an orthogonal basis $\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_r\}$ for $W$ that can be normalized to form an orthonormal basis.

Let ${𝐯}_1={𝐮}_1$ and let $W_1$ be the space spanned by ${𝐯}_1$.

The line $W_1$ contains ${𝐯}_1={𝐮}_1$. The vector ${𝐮}_2$ points up and to the right. Its projection ${\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_1}{𝐮}_2$ lies along $W_1$, and ${𝐯}_2={𝐮}_2-{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_1}{𝐮}_2$ starts at the origin and is perpendicular to $W_1$.

Let

$${\displaystyle {𝐯}_2={𝐮}_2-{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_1}{𝐮}_2={𝐮}_2-\frac{⟨{𝐮}_2,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1}$$

We know ${𝐯}_2\ne \mathrm{𝟎}$, because if it were, ${𝐮}_2$ would be a scalar multiple of ${𝐯}_1={𝐮}_1$, contradicting that $\{{𝐮}_1,\dots ,{𝐮}_r\}$ is linearly independent. By construction, ${𝐯}_1⟂{𝐯}_2$.

Now let $W_2$ be the space spanned by ${𝐯}_1$ and ${𝐯}_2$.

A shaded plane $W_2$ contains ${𝐯}_1$ and ${𝐯}_2$. The vector ${𝐮}_3$ points up out of the plane. Its projection ${\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_2}{𝐮}_3$ lies in the plane, directly below the tip of ${𝐮}_3$, and ${𝐯}_3={𝐮}_3-{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_2}{𝐮}_3$ starts at the origin and is perpendicular to the plane.

$${\displaystyle {𝐯}_3={𝐮}_3-{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_2}{𝐮}_3={𝐮}_3-\frac{⟨{𝐮}_3,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1-\frac{⟨{𝐮}_3,{𝐯}_2⟩}{\Vert {𝐯}_2{\Vert }^2}{𝐯}_2}$$

is orthogonal to $W_2$, and so is orthogonal to ${𝐯}_1$ and ${𝐯}_2$. We can likewise find a vector ${𝐯}_4$ that is orthogonal to ${𝐯}_1$, ${𝐯}_2$, and ${𝐯}_3$:

$${\displaystyle {𝐯}_4={𝐮}_4-\frac{⟨{𝐮}_4,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1-\frac{⟨{𝐮}_4,{𝐯}_2⟩}{\Vert {𝐯}_2{\Vert }^2}{𝐯}_2-\frac{⟨{𝐮}_4,{𝐯}_3⟩}{\Vert {𝐯}_3{\Vert }^2}{𝐯}_3}$$

and so on until we have $r$ vectors. Normalizing the ${𝐯}_i$ results in an orthonormal basis.

Example

Let $R^4$ have the Euclidean inner product. Use the Gram-Schmidt process to transform the basis $\{{𝐮}_1,{𝐮}_2,{𝐮}_3,{𝐮}_4\}$ into an orthonormal basis.

${𝐮}_1=(1,1,0,0)$, ${𝐮}_2=(1,0,1,0)$, ${𝐮}_3=(0,1,1,1)$, ${𝐮}_4=(-1,1,1,1)$

Solution

We want an orthogonal set $\{{𝐯}_1,{𝐯}_2,{𝐯}_3,{𝐯}_4\}$ first, then we normalize. Start with

$${𝐯}_1={𝐮}_1=(1,1,0,0)$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐯}_2}& {\displaystyle ={𝐮}_2-\frac{{𝐮}_2·{𝐯}_1}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1=(1,0,1,0)-\frac{1}{2}(1,1,0,0)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(\frac{1}{2},-\frac{1}{2},1,0)}\hfill \end{array}}$$

Scaling doesn’t affect orthogonality, so use ${𝐯}_2\prime =2{𝐯}_2=(1,-1,2,0)$ to avoid fractions.

Tip: Check for orthogonality as you go: ${𝐯}_1·{𝐯}_2\prime =1-1+0+0=0$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐯}_3}& {\displaystyle ={𝐮}_3-\frac{{𝐮}_3·{𝐯}_1}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1-\frac{{𝐮}_3·{𝐯}_2\prime }{\Vert {𝐯}_2\prime {\Vert }^2}{𝐯}_2\prime }\hfill \\ \hfill {\displaystyle }& {\displaystyle =(0,1,1,1)-\frac{1}{2}(1,1,0,0)-\frac{1}{6}(1,-1,2,0)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-\frac{2}{3},\frac{2}{3},\frac{2}{3},1)}\hfill \end{array}}$$

Use ${𝐯}_3\prime =3{𝐯}_3=(-2,2,2,3)$.

Tip: Check: ${𝐯}_1·{𝐯}_3\prime =-2+2=0$ and ${𝐯}_2\prime ·{𝐯}_3\prime =-2-2+4=0$.

Here ${𝐮}_4·{𝐯}_1=0$ and ${𝐮}_4·{𝐯}_2\prime =-1-1+2=0$, so those terms drop out, and ${𝐮}_4·{𝐯}_3\prime =2+2+2+3=9$ with $\Vert {𝐯}_3\prime {\Vert }^2=21$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐯}_4}& {\displaystyle ={𝐮}_4-\frac{{𝐮}_4·{𝐯}_1}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1-\frac{{𝐮}_4·{𝐯}_2\prime }{\Vert {𝐯}_2\prime {\Vert }^2}{𝐯}_2\prime -\frac{{𝐮}_4·{𝐯}_3\prime }{\Vert {𝐯}_3\prime {\Vert }^2}{𝐯}_3\prime }\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-1,1,1,1)-0-0-\frac{3}{7}(-2,2,2,3)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(-\frac{1}{7},\frac{1}{7},\frac{1}{7},-\frac{2}{7})}\hfill \end{array}}$$

Use ${𝐯}_4\prime =7{𝐯}_4=(-1,1,1,-2)$.

Tip: Check: ${𝐯}_4\prime ·{𝐯}_1=0$, ${𝐯}_4\prime ·{𝐯}_2\prime =0$, ${𝐯}_4\prime ·{𝐯}_3\prime =0$.

Orthogonal basis:

$$\{(1,1,0,0),(1,-1,2,0),(-2,2,2,3),(-1,1,1,-2)\}$$

The norms are $\sqrt{2}$, $\sqrt{6}$, $\sqrt{21}$, and $\sqrt{7}$. Dividing each vector by its norm gives the orthonormal basis

Theorem 6.3.6 (inner product space analog of Theorem 4.6.5 (b))

Example 6.3.9 Legendre Polynomials

Let the vector space $P_2$ have the inner product ${\displaystyle ⟨𝐩,𝐪⟩={\int }_{-1}^{1}p(x)q(x)dx}$. Apply the Gram-Schmidt process to transform the standard basis $\{1,x,x^2\}$ for $P_2$ into an orthogonal basis $\{{\varphi }_1(x),{\varphi }_2(x),{\varphi }_3(x)\}$.

Solution

Start with ${𝐮}_1=1$, ${𝐮}_2=x$, ${𝐮}_3=x^2$.

$${\varphi }_1(x)=1$$

$${\displaystyle {\varphi }_2(x)=x-\frac{⟨x,1⟩}{\Vert 1{\Vert }^2}·1,\text{where}⟨x,1⟩={\int }_{-1}^{1}xdx=0}$$

$${\varphi }_2(x)=x$$

$${\displaystyle {\varphi }_3(x)=x^2-\frac{⟨x^2,1⟩}{\Vert 1{\Vert }^2}·1-\frac{⟨x^2,x⟩}{\Vert x{\Vert }^2}x}$$

Compute the pieces:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ⟨x^2,x⟩}& {\displaystyle ={\int }_{-1}^1{x}^{3}dx=0}\hfill \\ \hfill {\displaystyle ⟨x^2,1⟩}& {\displaystyle ={\int }_{-1}^1{x}^{2}dx={\frac{1}{3}{x}^3|}_{-1}^1=\frac{2}{3}}\hfill \\ \hfill {\displaystyle \Vert 1{\Vert }^2}& {\displaystyle ={\int }_{-1}^{1}1dx=2}\hfill \end{array}}$$

$${\displaystyle {\varphi }_3(x)=x^2-\frac{2/3}{2}=x^2-\frac{1}{3}}$$

Suppose we want a scalar $a$ such that $a{\varphi }_3(1)=1$. Then

$${\displaystyle a(1-\frac{1}{3})=1\Rightarrow \frac{2}{3}a=1\Rightarrow a=\frac{3}{2},a{\varphi }_3=\frac{3}{2}(x^2-\frac{1}{3})=\frac{1}{2}(3x^2-1)}$$

We then have ${\displaystyle \{1,x,\frac{1}{2}(3x^2-1)\}}$.

Definition

The first three Legendre polynomials are $1$, $x$, and ${\displaystyle \frac{1}{2}(3x^2-1)}$. They result from transforming the standard basis for $P_2$ with the inner product ${\displaystyle ⟨𝐩,𝐪⟩={\int }_{-1}^{1}p(x)q(x)dx}$ into an orthogonal basis for $P_2$ via the Gram-Schmidt process, scaled so they have a value of 1 at $x=1$.