- #1

joe_cool2

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## Homework Statement

Let R

^{2}have the Euclidean inner product and use the Gram-Schmidt process to transform the basis {u

_{1},u

_{2}} into an orthonormal basis.

u

_{1}= (1,-3)

u

_{2}= (2,2)

## Homework Equations

Gram-Schmidt process:

[tex]

\\v_1 = u_1

\\v_2= u_2 -

\frac{\left ( \left \langle u_2, v_1\right \rangle \right )}{\left \|

v_1

\right \| ^ 2}v_1

[/tex]

## The Attempt at a Solution

We'll go ahead and find the norm of v first. Then we'll evaluate the second Gram-Schmidt equation.

[tex]

\\\left \| v_1 \right \| = \sqrt{10} \\\\

\left \| v_1 \right \|v_1 = u_1 = (1,-3)

\\\\\left \| v_2 \right \|v_2= \begin{bmatrix}

2\\2

\end{bmatrix} -

\frac{\left ( \left \langle \begin{bmatrix}

2\\2

\end{bmatrix} , \begin{bmatrix}

\frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}

\end{bmatrix} \right \rangle \right )}{\left \| \begin{bmatrix}

\frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}

\end{bmatrix} \right \| ^ 2}

\begin{bmatrix}

\frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}}

\end{bmatrix}\\\\\left \| v_2 \right \|v_2= \begin{bmatrix}

2\\2

\end{bmatrix} - \frac{\left ( \frac{-4}{\sqrt{10}} \right )}{1} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix}

\\\\

\begin{bmatrix}

2\\2

\end{bmatrix}

- \begin{bmatrix}

\frac{-4}{10}\\\\\frac{12}{10}

\end{bmatrix}

=

\begin{bmatrix}

\frac{24}{10}\\\\\frac{8}{10}

\end{bmatrix}

=

\begin{bmatrix}

\frac{12}{5}\\\\\frac{4}{5}

\end{bmatrix}

[/tex]

I've not been able to get another answer for this step. Even checked in Matlab.

Now, we'll normalize the vector like the problem wants us to:

[tex]

\\\left \| v_2 \right \| = \sqrt{8}

\\

v_2 = \begin{bmatrix}

\frac{\frac{12}{5}}{\sqrt{8}}\\

\frac{\frac{4}{5}}{\sqrt{8}}

\end{bmatrix}

=

\begin{bmatrix}

\frac{\frac{12}{5}}{\sqrt{8}}\\

\frac{\frac{4}{5}}{\sqrt{8}}

\end{bmatrix}

=

\begin{bmatrix}

\frac{12}{10\sqrt{2}}\\\\

\frac{4}{10\sqrt{2}}

\end{bmatrix}

=

\begin{bmatrix}

\frac{6}{5\sqrt{2}}\\\\

\frac{2}{5\sqrt{2}}

\end{bmatrix}

=

\begin{bmatrix}

\frac{6\sqrt{2}}{10}\\\\

\frac{2\sqrt{2}}{10}

\end{bmatrix}

=

\begin{bmatrix}

\frac{3\sqrt{2}}{5}\\\\

\frac{\sqrt{2}}{5}

\end{bmatrix}

[/tex]

This is an odd problem, so I know that the solution should be [itex] v_2 = (\frac{3}{\sqrt{10}} , \frac{1}{\sqrt{10}}) [/itex]. There must be something wrong when I'm finding [itex]||v_2||v_2[/itex] . Is there anything fundamentally wrong with my process?

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