# Orthogonal set

• Nov 5th 2007, 11:25 PM
namelessguy
Orthogonal set
Can anyone help me with this problem please?
Let {v_1,v_2,...v_k} be an orthogonal set in V, and let a_1,a_2,...,a_k be scalars. Prove that (the norm of the summation of a_i*v_i)^2 is equal to the summation of l a_i l^2* (norm of v_i)^2
i runs from 1 to k.
• Nov 6th 2007, 03:57 AM
kalagota
Quote:

Originally Posted by namelessguy
Can anyone help me with this problem please?
Let {v_1,v_2,...v_k} be an orthogonal set in V, and let a_1,a_2,...,a_k be scalars. Prove that (the norm of the summation of a_i*v_i)^2 is equal to the summation of l a_i l^2* (norm of v_i)^2
i runs from 1 to k.

${\left| \left|{ \sum_{i=1}^k {a_i v_i}} \right| \right|}^2 = \left<{{ \sum_{i=1}^k {a_i v_i}},{ \sum_{i=1}^k {a_i v_i}}}\right> = \sum_{i=1}^k\left<{a_i v_i,a_i v_i}\right>$

$= \sum_{i=1}^k {a_i}^2\left<{v_i,v_i}\right> = \sum_{i=1}^k {a_i}^2\| v_i \| ^2$

im just trying my luck here.. please, someone verify my proof.. thx..
• Nov 6th 2007, 07:55 AM
ThePerfectHacker
Quote:

Originally Posted by kalagota
im just trying my luck here.. please, someone verify my proof.. thx..

It looks right.
• Nov 6th 2007, 10:25 PM
namelessguy
Thanks a lot for your help, kalagota. Where can I learn to type mathematical symbols and does it take a lot of time to learn this?
• Nov 6th 2007, 11:57 PM
Opalg
Quote:

Originally Posted by kalagota
${\left| \left|{ \sum_{i=1}^k {a_i v_i}} \right| \right|}^2 = \left<{{ \sum_{i=1}^k {a_i v_i}},{ \sum_{i=1}^k {a_i v_i}}}\right> = \sum_{i=1}^k\left<{a_i v_i,a_i v_i}\right>$

This should be $\Bigl\|\sum_{i=1}^k a_i v_i \Bigr\|^2 = \Bigl\langle\sum_{i=1}^k a_i v_i, \sum_{j=1}^k a_j v_j\Bigr\rangle = \sum_{i,j=1}^k \langle a_i v_i,a_j v_j\rangle$. The two sums are independent, so should have different dummy variables. However, $\langle v_i,v_j\rangle = 0$ when i≠j. So the only terms that survive are those for which j=i, and the rest of kalagota's solution is correct:

Quote:

Originally Posted by kalagota
$= \sum_{i=1}^k {a_i}^2\left<{v_i,v_i}\right> = \sum_{i=1}^k {a_i}^2\| v_i \| ^2$

The only other comment is that if the scalar field is the complex numbers then when $a_i$ comes out of the right side of the inner product it becomes $\bar{a}_i$ (the complex conjugate). But that's okay, because we then get $a_i\bar{a}_i=|a_i|^2$, as required.
• Nov 7th 2007, 03:26 AM
kalagota
Quote:

Originally Posted by Opalg
This should be $\Bigl\|\sum_{i=1}^k a_i v_i \Bigr\|^2 = \Bigl\langle\sum_{i=1}^k a_i v_i, \sum_{j=1}^k a_j v_j\Bigr\rangle = \sum_{i,j=1}^k \langle a_i v_i,a_j v_j\rangle$. The two sums are independent, so should have different dummy variables. However, $\langle v_i,v_j\rangle = 0$ when i≠j. So the only terms that survive are those for which j=i, and the rest of kalagota's solution is correct:
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