# Quiz 10. Find Orthogonal Basis / Find Value of Linear Transformation ## Problem 356

(a) Let $S=\{\mathbf{v}_1, \mathbf{v}_2\}$ be the set of the following vectors in $\R^4$.
$\mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}.$ Find an orthogonal basis of the subspace $\Span(S)$ of $\R^4$.

(b) Let $T:\R^2 \to \R^3$ be a linear transformation such that
$T(\mathbf{e}_1)=\mathbf{u}_1 \text{ and } T(\mathbf{e}_2)=\mathbf{u}_2,$ where $\{\mathbf{e}_1, \mathbf{e}_2\}$ is the standard unit vectors of $\R^2$ and
$\mathbf{u}_1=\begin{bmatrix} 5 \\ 1 \\ 2 \end{bmatrix} \text{ and } \mathbf{u}_2=\begin{bmatrix} 8 \\ 2 \\ 6 \end{bmatrix}.$ Then find
$T\left(\, \begin{bmatrix} 3 \\ -2 \end{bmatrix} \,\right).$ Add to solve later

## (a) Solution 1. (Using the Gram-Schmidt process)

It is straightforward to check that the vectors $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, and hence the set $S$ is a basis of $\Span(S)$.
Since the dot (inner) product of $\mathbf{v}_1$ and $\mathbf{v}_2$ is
$\mathbf{v_1}\cdot \mathbf{v}_2=1\neq 0,$ $S$ is not an orthogonal basis. We apply the Gram-Schmidt process to generate an orthogonal basis from the basis $S$.

The Gram-Schmidt process for two vectors is as follows. We define vectors $\mathbf{u}_1, \mathbf{u}_2$ by the following formula. Then $B=\{\mathbf{u}_1, \mathbf{u_2} \}$ is an orthogonal basis of $\Span(S)$.
\begin{align*}
\mathbf{u}_1&:=\mathbf{v}_1\6pt] \mathbf{u}_2&:=\mathbf{v}_2-\frac{\mathbf{u}_1\cdot \mathbf{v}_2}{\mathbf{u}_1\cdot \mathbf{u}_1} \mathbf{u}_1. \tag{*} \end{align*} Since we have \begin{align*} \mathbf{u}_1 \cdot \mathbf{v}_2=\mathbf{v}_1\cdot \mathbf{v}_2=1 \text{ and } \mathbf{u}_1\cdot \mathbf{u}_1=\mathbf{v}_1\cdot \mathbf{v}_1=2, \end{align*} we compute \begin{align*} \mathbf{u}_2&=\mathbf{v}_2-\frac{1}{2}\mathbf{u}_1\\[6pt] &=\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}-\frac{1}{2} \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}\\[6pt] &=\begin{bmatrix} -1/2\\ 1 \\ 1/2 \\ 0 \end{bmatrix}=\frac{1}{2}\begin{bmatrix} -1 \\ 2 \\ 1 \\ 0 \end{bmatrix}. \end{align*} Therefore the set \[\left\{\, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1/2\\ 1 \\ 1/2 \\ 0 \end{bmatrix}\,\right\} is an orthogonal basis of $\Span(S)$.
Note that scaling by a nonzero scalar does not change the orthogonality, the set
$\left\{\, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1\\ 2 \\ 1 \\ 0 \end{bmatrix}\,\right\}$ is also an orthogonal basis of $\Span(S)$, just in case you prefer not to have a fraction.

## (a) Solution 2. (Using a pattern of the Gram-Schmidt process)

Here is another solution using a partial information of the Gram-Schmidt process.
As in Solution 1, the set $S$ is a (non-orthogonal) basis of $\Span(S)$.
We want to apply the Gram-Schmidt process but suppose we only remember the pattern of the Gram-Schmidt process. Namely, we want to define orthogonal vectors $\mathbf{u}, \mathbf{u}_2$ by
\begin{align*}
\mathbf{u}_1&:=\mathbf{v}_1\6pt] \mathbf{u}_2&:=\mathbf{v}_2+a \mathbf{u}_1. \end{align*} Here a is some number, which is given in the Gram-Schmidt process (*) but we don’t remember. We can still determine the number a as follows. Since \mathbf{u}_1 and \mathbf{u_2} will be orthogonal, we have \begin{align*} 0&=\mathbf{u}_1\cdot \mathbf{u}_2=\mathbf{u}_1 \cdot (\mathbf{v}_2+a\mathbf{u}_1)\\ &=\mathbf{u}_1\cdot \mathbf{v}_2+a\mathbf{u}_1 \cdot \mathbf{u}_1\\ &=1+2a. \end{align*} Hence, we obtain a=-1/2. Then we determine \begin{align*} \mathbf{u}_2&=\mathbf{v}_2-\frac{1}{2}\mathbf{u}_1\\[6pt] &=\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}-\frac{1}{2} \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}\\[6pt] &=\begin{bmatrix} -1/2\\ 1 \\ 1/2 \\ 0 \end{bmatrix}=\frac{1}{2}\begin{bmatrix} -1 \\ 2 \\ 1 \\ 0 \end{bmatrix}. \end{align*} (So we could complete the Gram-Schmidt process even though we didn’t remember the details.) Hence the set \[\left\{\, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1/2\\ 1 \\ 1/2 \\ 0 \end{bmatrix}\,\right\} is an orthogonal basis of $\Span(S)$.

## (b) Solution.

We first express the vector $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$ as the linear combination
$\begin{bmatrix} 3 \\ -2 \end{bmatrix}=3\begin{bmatrix} 1 \\ 0 \end{bmatrix}-2\begin{bmatrix} 0 \\ 1 \end{bmatrix}=3\mathbf{e}_1-2\mathbf{e}_2.$ Then we compute
\begin{align*}
T\left(\, \begin{bmatrix}
3 \\
-2
\end{bmatrix} \,\right)&=T(3\mathbf{e}_1-2\mathbf{e}_2)\\
&=3T(\mathbf{e}_1)-2T(\mathbf{e}_2) && \text{ by linearity of $T$}\6pt] &=3\begin{bmatrix} 5 \\ 1 \\ 2 \end{bmatrix}-2\begin{bmatrix} 8 \\ 2 \\ 6 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 15 \\ 3 \\ 6 \end{bmatrix}-\begin{bmatrix} 16 \\ 4 \\ 12 \end{bmatrix}=\begin{bmatrix} -1 \\ -1 \\ -6 \end{bmatrix}. \end{align*} Therefore we have found \[T\left(\, \begin{bmatrix} 3 \\ -2 \end{bmatrix} \,\right) =\begin{bmatrix} -1 \\ -1 \\ -6 \end{bmatrix}.

## Comment.

These are Quiz 10 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. Add to solve later

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