# Linear Algebra - Gram-Schmidt Process

## Homework Statement

Let R2 have the Euclidean inner product and use the Gram-Schmidt process to transform the basis {u1,u2} into an orthonormal basis.

u1 = (1,-3)
u2 = (2,2)

## Homework Equations

Gram-Schmidt process:

$$\\v_1 = u_1 \\v_2= u_2 - \frac{\left ( \left \langle u_2, v_1\right \rangle \right )}{\left \| v_1 \right \| ^ 2} v_1$$

## 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.

$$\\\left \| v_1 \right \| = \sqrt{10} \\\\ \left \| v_1 \right \|v_1 = u_1 = (1,-3) \\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \left \langle \begin{bmatrix} 2\\2 \end{bmatrix} , \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \rangle \right )}{\left \| \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \| ^ 2} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \frac{-4}{\sqrt{10}} \right )}{1} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \\\\ \begin{bmatrix} 2\\2 \end{bmatrix} - \begin{bmatrix} \frac{-4}{10}\\\\\frac{12}{10} \end{bmatrix} = \begin{bmatrix} \frac{24}{10}\\\\\frac{8}{10} \end{bmatrix} = \begin{bmatrix} \frac{12}{5}\\\\\frac{4}{5} \end{bmatrix}$$

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:
$$\\\left \| v_2 \right \| = \sqrt{8} \\ v_2 = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{12}{10\sqrt{2}}\\\\ \frac{4}{10\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6}{5\sqrt{2}}\\\\ \frac{2}{5\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6\sqrt{2}}{10}\\\\ \frac{2\sqrt{2}}{10} \end{bmatrix} = \begin{bmatrix} \frac{3\sqrt{2}}{5}\\\\ \frac{\sqrt{2}}{5} \end{bmatrix}$$

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

Last edited:

vela
Staff Emeritus
Homework Helper

## Homework Statement

Let R2 have the Euclidean inner product and use the Gram-Schmidt process to transform the basis {u1,u2} into an orthonormal basis.

u1 = (1,-3)
u2 = (2,2)

## Homework Equations

Gram-Schmidt process:

$$\\v_1 = u_1 \\v_2= u_2 - \frac{\left ( \left \langle u_2, v_1\right \rangle \right )}{\left \| v_1 \right \| ^ 2} v_1$$

## 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.

$$\\\left \| v_1 \right \| = \sqrt{10} \\\\ \left \| v_1 \right \|v_1 = u_1 = (1,-3) \\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \left \langle \begin{bmatrix} 2\\2 \end{bmatrix} , \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \rangle \right )}{\left \| \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \right \| ^ 2} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \\\\\left \| v_2 \right \|v_2= \begin{bmatrix} 2\\2 \end{bmatrix} - \frac{\left ( \frac{-4}{\sqrt{10}} \right )}{1} \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{-3}{\sqrt{10}} \end{bmatrix} \\\\ \begin{bmatrix} 2\\2 \end{bmatrix} - \begin{bmatrix} \frac{-4}{10}\\\\\frac{12}{10} \end{bmatrix} = \begin{bmatrix} \frac{24}{10}\\\\\frac{8}{10} \end{bmatrix} = \begin{bmatrix} \frac{12}{5}\\\\\frac{4}{5} \end{bmatrix}$$

I've not been able to get another answer for this step. Even checked in Matlab.
If you get rid of a common factor of 4/5, you have ##\vec{v}_2 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}##.

Now, we'll normalize the vector like the problem wants us to:
$$\\\left \| v_2 \right \| = \sqrt{8} \\ v_2 = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{\frac{12}{5}}{\sqrt{8}}\\ \frac{\frac{4}{5}}{\sqrt{8}} \end{bmatrix} = \begin{bmatrix} \frac{12}{10\sqrt{2}}\\\\ \frac{4}{10\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6}{5\sqrt{2}}\\\\ \frac{2}{5\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{6\sqrt{2}}{10}\\\\ \frac{2\sqrt{2}}{10} \end{bmatrix} = \begin{bmatrix} \frac{3\sqrt{2}}{5}\\\\ \frac{\sqrt{2}}{5} \end{bmatrix}$$

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

Ok. So I gather that my answer is not technically wrong? How can I verify that my original answer is in fact an orthonormal basis spanning the same subspace?

I guess if I make computational errors, it will be unlikely that all the dot products in some generated set of vectors is zero. Right?

Last edited:
vela
Staff Emeritus