# Orthonormal Vectors

## Homework Statement

http://img130.imageshack.us/img130/2217/49541511.gif [Broken]

## The Attempt at a Solution

I'm not sure how I need to start but I think I need to use the Gram-Schmidt orthogonalization process to first to transform this into an orthogonal basis, and then normalize them to obtain an orthonormal basis. So we have:

$$w_1=(1,1,1,0)^T$$
$$w_2=(0,1,1,1)^T$$

$$v_1=w_1=(1,1,1,0)^T$$

$$v_2=w_2 - proj_{w_1}w_2 = (1,1,1,0)^T - \frac{w_2 . v_1}{\| v_1 \|}v_1$$

$$= \left(\begin{array}{ccc}0\\1\\1\\1\end{array}\right) - \frac{\left(\begin{array}{ccc}0\\1\\1\\1\end{array}\right).\left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right)}{\left\| \left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right) \right\|} \left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right)$$

$$= \left(\begin{array}{ccc}0\\1\\1\\1\end{array}\right)- \frac{2}{3} \left(\begin{array}{ccc}1\\1\\1\\0\end{array}\right) = \left(\begin{array}{ccc}-2/3\\1/3\\1/3\\1/3\end{array}\right)$$

Now I normalize the the vectors:

$$\left\| v_1 \right\| = \sqrt{3}$$

$$\left\| v_2 \right\| = \sqrt{\left(\frac{-2}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2} = \sqrt{\frac{7}{9}}$$

$$q_1 = \frac{v_1}{\|v_1 \|} = \left(\begin{array}{ccc}1/\sqrt{3}\\1/\sqrt{3}\\1/\sqrt{3}\\0\end{array}\right)$$

$$q_2 = \frac{v_2}{\|v_2 \|} = \left(\begin{array}{ccc}-2/3 \div \sqrt{7/9}\\1/3 \div \sqrt{7/9}\\1/3 \div \sqrt{7/9}\\1/3\div \sqrt{7/9}\end{array}\right)$$

I don't know if my method and my results are correct. Is there anything wrong with my working?

Last edited by a moderator:

vela
Staff Emeritus
Homework Helper
Your formula for v2 should say $\| v_1 \|^2$ on the bottom. That's actually how you did the calculation, though, so your answer is correct. You can simplify v2 by pulling the 9 out of the square root and canceling it with the 3 in the numerator.

You can always check your answer by finding the dot product of v1 and v2 and making sure it's zero, and double-checking to make sure the two vectors are normalized.

jbunniii
Homework Helper
Gold Member
I don't think your answer is right. If the Gram-Schmidt procedure is carried out correctly, $\{q_1,q_2\}$ and $\{w_1,w_2\}$ should be bases for the same subspace. In particular, it should be possible to express $w_2$ as a linear combination of $q_1$ and $q_2$. It's not hard to see that this is impossible with your $q_1$ and $q_2$.

vela
Staff Emeritus
Homework Helper
jbunniii is right. I missed the arithmetic mistake you made when calculating the fourth component of v2. Your general method is correct, however, even if your execution of it wasn't perfect.

Thanks you everyone.

So,

$$v_2 = \left(\begin{array}{ccc}-2/3\\1/3\\1/3\\1\end{array}\right)$$

Therefore

$$q_2 = \frac{v_2}{\|v_2 \|} = \left(\begin{array}{ccc}-2/3 \div \sqrt{7/9}\\1/3 \div \sqrt{7/9}\\1/3 \div \sqrt{7/9}\\1/ \sqrt{7/9}\end{array}\right)$$

About the simplification, I'm not sure if I understand this properly:

You can simplify v2 by pulling the 9 out of the square root and canceling it with the 3 in the numerator.

How can I pull the 9 out of the square root?

So, how should I simplify q2?

jbunniii
Homework Helper
Gold Member
So, how should I simplify q2?

As vela said, you can pull the 9 out of the square root since it's equal to $3^2$:

$$\sqrt{\frac{7}{9}} = \frac{1}{3}\sqrt{7}$$

$$\sqrt{\frac{7}{9}}$$

it's more conventional to MULTIPLY by the reciprocal, which is

$$\sqrt{\frac{9}{7}} = \frac{3}{\sqrt{7}}$$

and if you don't like square roots in the denominator (I personally think they're A-OK), this is equivalent to

$$\frac{3\sqrt{7}}{7}$$