Orthonormal Vectors Homework: Solution

[roam]

Homework Statement

http://img130.imageshack.us/img130/2217/49541511.gif

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?

 

[vela]

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]

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]

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.

 

[roam]

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?

 

[roam]

So, how should I simplify q₂?

 

[jbunniii]

So, how should I simplify q₂?

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

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

Also, instead of DIVIDING by

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

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

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