1. ## 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.

2. 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..

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

4. 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?

5. 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:

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.

6. 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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