how to find a basis for a set of vectors

Problem 600

Permit $\mathbf{v}_1=\brainstorm{bmatrix}
2/three \\ 2/3 \\ 1/iii
\end{bmatrix}$ be a vector in $\R^3$.

Observe an orthonormal basis for $\R^3$ containing the vector $\mathbf{v}_1$.

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Solution ane (The Gram-Schumidt Orthogonalization)

Outset of all, note that the length of the vector $\mathbf{v}_1$ is $1$ as
\[\|\mathbf{v}_1\|=\sqrt{\left(\frac{two}{3}\right)^2+\left(\frac{2}{3}\right)^2+\left(\frac{one}{3}\right)^2}=i.\]

We want to detect two vectors $\mathbf{five}_2, \mathbf{v}_3$ such that $\{\mathbf{5}_1, \mathbf{v}_2, \mathbf{five}_3\}$ is an orthonormal basis for $\R^3$.
The vectors $\mathbf{five}_2, \mathbf{5}_3$ must lie on the aeroplane that is perpendicular to the vector $\mathbf{v}_1$.
So consider the subspace
\[Due west=\left \{\,\brainstorm{bmatrix} ten\\y\\z \cease{bmatrix} \in \R^3 \quad \middle | \quad \begin{bmatrix} x\\y\\z \end{bmatrix} \cdot \begin{bmatrix}
ii/three \\ ii/three \\ 1/3
\end{bmatrix}=0 \,\right\}\]

Note that $W$ consists of all vectors that are perpendicular to $\mathbf{v}_1$, hence $W$ is a plane that is perpendicular to $\mathbf{v}_1$.

The relation
\[ \begin{bmatrix} x\\y\\z \end{bmatrix} \cdot \begin{bmatrix}
2/3 \\ 2/3 \\ 1/3
\end{bmatrix}=0\] tin can be written as
\[\frac{2}{3}10+\frac{two}{3}y+\frac{1}{3}z=0,\] or equivalently
\[z=-2x-2y.\] Hence the vectors in $West$ can be written every bit
\[\begin{bmatrix} x\\y\\z \stop{bmatrix}=\begin{bmatrix} x\\y\\-2x-2y \terminate{bmatrix}=ten\begin{bmatrix}i\\0\\-2\end{bmatrix}+y\brainstorm{bmatrix}0\\one\\-2\end{bmatrix}.\]

It follows that
\[\left\{\,\begin{bmatrix}one\\0\\-2\end{bmatrix}, \begin{bmatrix}0\\i\\-two\end{bmatrix} \,\right\}\] is a basis for the subspace $W$. Let united states call these vectors $\mathbf{u}_1, \mathbf{u}_2$, respectively.

Nosotros utilize the Gram-Schmidt orthogonalization to this ground $\{\mathbf{u}_1, \mathbf{u}_2\}$ and obtain an orthogonal basis every bit follows.
We do non modify the first vector: let $\mathbf{w}_1=\mathbf{u}_1$.

Next, we set
\[\mathbf{w}_2=\mathbf{u}_2+a\mathbf{u}_1\] for some scalar $a$.
To determine $a$, we compute
\brainstorm{align*}
0&=\mathbf{westward}_1\cdot \mathbf{w}_2=\mathbf{u}_1\cdot(\mathbf{u}_2+a\mathbf{u}_1) \\
&=\mathbf{u}_1\cdot\mathbf{u}_2+a\mathbf{u}_1\cdot\mathbf{u}_1\\
&=(1\cdot 0+0\cdot one+(-2)\cdot(-2))+a(1\cdot i+ 0\cdot 0 +(-two)\cdot (-2)\\
&=4+5a.
\end{align*}
Hence, $a=-4/5$ and we obtain
\begin{marshal*}
\mathbf{westward}_2&=\mathbf{u}_2-\frac{4}{5}\mathbf{u}_1\\[6pt] &=\begin{bmatrix}0\\1\\-two\stop{bmatrix} -\frac{4}{v}\brainstorm{bmatrix}ane\\0\\-2\terminate{bmatrix}.
\end{marshal*}
(Note that you lot may utilize the Gram-Schumidt orthogonalization formula, instead of the higher up method.)

As scaling does not change the orthogonality, consider $5\mathbf{w}_2$, instead of $\mathbf{westward}_2$ (to avoids fractions).
We accept
\[5\mathbf{westward}_2=\begin{bmatrix}0\\five\\-x\end{bmatrix} -4\begin{bmatrix}i\\0\\-2\stop{bmatrix}=\begin{bmatrix}
-4\\ five\\-2
\terminate{bmatrix}.\]

Therefore, $\{\mathbf{w}_1, v\mathbf{w}_2\}$ is an orthogonal basis of $W$.

We obtain an orthonormal basis of $W$ by normalizing the length of these ground vectors.
Equally
\[\|\mathbf{w}_1\|=\sqrt{1^2+0^2+(-ii)^2}=\sqrt{five}\] and
\[\|5\mathbf{w}_2\|=\sqrt{(-four)^two+5^2+(-2)^2}=\sqrt{45}=3\sqrt{5},\] the vectors
\[\mathbf{v}_2:=\frac{\mathbf{w}_1}{\|\mathbf{westward}_1\|}=\frac{i}{\sqrt{5}}\brainstorm{bmatrix}1\\0\\-two\end{bmatrix}\] and
\[\mathbf{v}_3:=\frac{5\mathbf{w}_2}{\|5\mathbf{w}_2\|}=\frac{1}{3\sqrt{5}}\begin{bmatrix}-4\\v\\-two\end{bmatrix}\] form an orthonormal basis of $Due west$.
Note that as the vectors $\mathbf{v}_2, \mathbf{v}_3$ lie in $W$, they are still perpendicular to the vector $\mathbf{five}_1$.

Information technology follows that $\{\mathbf{v}_1, \mathbf{five}_2, \mathbf{v}_3\}$ is an orthonomal set in $\R^3$, thus information technology is an orthonormal basis for $\R^3$.

Solution 2 (Cross Product)

Next, we solve the trouble using the cantankerous product.

Let $\mathbf{v}_1=\frac{1}{3}\mathbf{u}_1$, where $\mathbf{u}_1=\begin{bmatrix}
2 \\
2 \\
ane
\end{bmatrix}$.
Our starting time goal is to detect the vectors $\mathbf{u}_2$ and $\mathbf{u}_3$ such that $\{\mathbf{u}_1,\mathbf{u}_2, \mathbf{u}_3\}$ is an orthogonal basis for $\R^3$.

Let $\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}$ exist a vector that is perpendicular to $\mathbf{u}_1$.
Then we take $\mathbf{10}\cdot \mathbf{u}_1=0$, and hence we have the relation
\[2x+2y+z=0.\] For example, the vector $\mathbf{u}_2:=\begin{bmatrix}
i \\
0 \\
-2
\end{bmatrix}$ satisfies the relation, and hence $\mathbf{u}_2\cdot \mathbf{u}_1=0$.
(So far, it is not so different from Solution 1.)

Now, let united states of america define the third vector $\mathbf{u}_3$ to exist the cross product of $\mathbf{u}_1$ and $\mathbf{u}_2$:
\[\mathbf{u}_3:=\mathbf{u}_1\times \mathbf{u}_2=\begin{bmatrix}
two \\
2 \\
ane
\end{bmatrix}\times \begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}=\begin{bmatrix}
-4 \\
5 \\
-2
\cease{bmatrix}.\] Past the property of the cantankerous production, the vector $\mathbf{u}_3$ is perpendicular to both $\mathbf{u}_1, \mathbf{u}_2$.

Therefore, the set
\[\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}=\left\{\,\begin{bmatrix}
two \\
2 \\
1
\end{bmatrix}, \begin{bmatrix}
ane \\
0 \\
-two
\end{bmatrix}, \begin{bmatrix}
-four \\
5 \\
-2
\end{bmatrix} \,\right\}\] is an orthogonal basis for $\R^3$ as it consists of three nonzero orthogonal vectors.

Finally, to obtain an orthonormal basis for $\R^2$, we simply need to normalize the lengths of these vectors.
The lengths are
\begin{align*}
\|\mathbf{u}_1\|&=\sqrt{2^two+2^2+1^2}=\sqrt{9}=3 \\
\|\mathbf{u}_2\|&=\sqrt{ane^2+0^2+(-2)^2}=\sqrt{v}\\
\|\mathbf{u}_3\|&=\sqrt{(-four)^two+5^2+(-two)^ii}=\sqrt{45}=3\sqrt{5}.
\end{marshal*}
Then we have $\mathbf{v}_1=\frac{\mathbf{u}_1}{\|\mathbf{u}_1\|}$.
Let $\mathbf{v}_2:=\frac{\mathbf{u}_2}{\|\mathbf{u}_2\|}$ and $\mathbf{five}_3:=\frac{\mathbf{u}_3}{\|\mathbf{u}_3\|}$.

Information technology follows that the set
\[\{\mathbf{v}_1\mathbf{v}_2\mathbf{v}_3\}=\left\{\, \frac{ane}{3}\begin{bmatrix}
2 \\
2 \\
1
\terminate{bmatrix}, \, \frac{1}{\sqrt{five}}\begin{bmatrix}ane\\0\\-ii\end{bmatrix}, \, \frac{1}{three\sqrt{5}}\begin{bmatrix}-4\\5\\-2\terminate{bmatrix} \,\right\}\] is an orthonormal footing for $\R^3$.

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