Let W=Span ({(i,0,1)}) in C^3. Find the orthonormal bases for W and W-perpendicular (W complement).

Can someone tell me what to do. I am lost.

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- Jul 22nd 2010, 06:01 AMivinewOrthnormal bases
Let W=Span ({(i,0,1)}) in C^3. Find the orthonormal bases for W and W-perpendicular (W complement).

Can someone tell me what to do. I am lost. - Jul 22nd 2010, 06:54 AMAckbeet
First step: find bases, not necessarily orthonormal, for those two spaces. Can you do that? What do you get?

- Jul 22nd 2010, 07:54 AMivinew
umm...can the standard basis be used for W like (1,0,0), (0,1,0), and (0,0,1)? I'm not sure how to find the bases for W perpendicular.

if not, then i should use the gram-schmidt process? if yes, then it'll be (i,0,1)? - Jul 22nd 2010, 07:57 AMAckbeet
Your candidate basis for W can't be correct, because it has three dimensions, whereas W only has one. I would just use the given vector as a basis. You can then normalize that vector (just divide by its length) to get an orthonormal basis for W. So much for W.

What about W-perp? W-perp consists of all the vectors that are perpendicular to the vectors in W. That means they'll all be perpendicular to the basis vector you've found in the previous paragraph. How could you characterize all those vectors? - Jul 22nd 2010, 10:08 AMivinew
OK, so when I normalize the vector do i take the result ,and then find the complement of that. For a complement to be perpendicular do we take an arbitrary vector (a,b,c) and find the inner product of our new found vector with the arbitrary vector and it has to equal zero?

W basis = ((radical 2)/2) (i,0,1)

So W complement = <W basis, (a,b,c)> =0 so (a,b,c) = (1, a, i) where a is any scalar in C? - Jul 22nd 2010, 10:25 AMAckbeet
Correct basis for W, I think. I think you're confusing yourself with respect to W-perp. You need (forget about the normalizing factor for now; it'll cancel out anyway):

$\displaystyle \langle(i,0,1),(a,b,c)\rangle=0,$ which implies

$\displaystyle ia+c=0.$

So you can see immediately that b can be anything, and that the first and third components must be related with this equation. Then how would you continue? - Jul 22nd 2010, 10:39 AMivinew
ok i put that in a matrix and solved the system of equations.

[i/radical 2 0 1/radical 2 |0] -> [i 0 1 | 0] -> [1 0 -i | 0 ] so x1 - ix3 =0

so x1 = ix3, x2= alpha, x3 = beta

beta (i,0,1) + alpha(0,1,0) - Jul 22nd 2010, 11:02 AMAckbeet
I'm not sure I can make out your method of solution, but your final answer makes sense to me. Here's a double-check: have you got two linearly independent vectors (all the vectors in a 3D space orthogonal to one vector should be a space of two dimensions)? Are both of those vectors orthogonal to the basis vector for W? Then you're good to go. I would, however, practice writing your method of solution in a clearer fashion! Communication skills are at least as important in the workplace as technical skills, if not more so.

Now you should orthonormalize those two vectors as basis vectors. I don't think you'll need to go through the whole Gram-Schmidt procedure, as they are already orthogonal, looks like. What do you get? - Jul 22nd 2010, 11:38 AMivinew
i put the two basis vectors together because dim (w) = 1 = dim (w perp). I got (i,1,1) and then i proceeded to normalize it and i get 1/rad 3 (i,1,1).

- Jul 22nd 2010, 11:41 AMAckbeet
Your post #9 makes no sense. You had beta (i,0,1) + alpha(0,1,0) for the basis of W-perp. That makes sense. It has dimension two. Don't second-guess yourself!

- Jul 22nd 2010, 11:46 AMivinew
Ooo ok.

so normalizing (i,0,1) = 1/rad 2 (i,0,1)

and normalizing (0,1,0) = (0,1,0) - Jul 22nd 2010, 11:52 AMAckbeet
Right. I think you're done!

- Jul 22nd 2010, 12:10 PMivinew
ooooo~ thank you so much for your quick reply!!!! ^_^

- Jul 22nd 2010, 12:34 PMAckbeet
You're very welcome. Have a good one!