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Orthonormal Vectors

  • Thread starter roam
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  • #1
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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:

[tex]w_1=(1,1,1,0)^T[/tex]
[tex]w_2=(0,1,1,1)^T[/tex]

[tex]v_1=w_1=(1,1,1,0)^T[/tex]

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

[tex]= \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)[/tex]

[tex]= \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)[/tex]

Now I normalize the the vectors:

[tex]\left\| v_1 \right\| = \sqrt{3}[/tex]

[tex]\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}}[/tex]

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

[tex]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)[/tex]

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

  • #2
vela
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Your formula for v2 should say [itex]\| v_1 \|^2[/itex] 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.
 
  • #3
jbunniii
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I don't think your answer is right. If the Gram-Schmidt procedure is carried out correctly, [itex]\{q_1,q_2\}[/itex] and [itex]\{w_1,w_2\}[/itex] should be bases for the same subspace. In particular, it should be possible to express [itex]w_2[/itex] as a linear combination of [itex]q_1[/itex] and [itex]q_2[/itex]. It's not hard to see that this is impossible with your [itex]q_1[/itex] and [itex]q_2[/itex].
 
  • #4
vela
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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.
 
  • #5
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Thanks you everyone. :smile:

So,

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

Therefore

[tex]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)[/tex]

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? :confused:
 
  • #6
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So, how should I simplify q2?
 
  • #7
jbunniii
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So, how should I simplify q2?
As vela said, you can pull the 9 out of the square root since it's equal to [itex]3^2[/itex]:

[tex]\sqrt{\frac{7}{9}} = \frac{1}{3}\sqrt{7}[/tex]

Also, instead of DIVIDING by

[tex]\sqrt{\frac{7}{9}}[/tex]

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

[tex]\sqrt{\frac{9}{7}} = \frac{3}{\sqrt{7}}[/tex]

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

[tex]\frac{3\sqrt{7}}{7}[/tex]
 

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