
#1
Nov1308, 09:41 PM

P: 70

1. The problem statement, all variables and given/known data
Find an orthonormal basis for the subspace of R^3 consisting of all vectors (a, b, c) such that a + b + c = 0. 2. Relevant equations 3. The attempt at a solution I know how to find an orthonormal basis just for R^3 by taking the standard basis vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1) and applying the GramSchmidt process to make them orthonormal. The condition that a + b + c must equal 0 is throwing me off, however. Can anyone give me any suggestions for how to approach problems like these? thanks! 



#2
Nov1308, 10:02 PM

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If you know GramSchmidt, that's the hard part. What's so hard about about finding two vectors (a,b,c) such that a+b+c=0 to do GramSchmidt with? (1,1,0) sounds like a good start. Give me another one.




#3
Nov1308, 10:07 PM

P: 70

I think I need 3 vectors, right? such that a + b + c = 0. Can I start with any vector? Or is there a reason why (1, 1, 0) is a good choice? If it can be any 3 vectors that add to 0 it should be simpler, but then I can't be sure they are a basis?




#4
Nov1308, 10:10 PM

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Finding an orthonormal basis
Why do you need 3 vectors? That's a subspace of R^3. It looks to me like it's two dimensional.




#5
Nov1308, 10:16 PM

P: 70

Ohh.. sorry I was being confused, I thought a, b, and c referred to vectors and not the components.
So I can choose (1, 1, 0) like you said and also (1, 0, 1)? I would still have to check if the 2 vectors I choose are actually a basis (span R^3 and are linearly independent) right? 



#6
Nov1308, 10:18 PM

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Sure, exactly. That's a good choice. There aren't too many dumb choices. (1,1,0) and (1,1,0) would have been a dumb choice because they're linearly dependent. But you didn't make the dumb choice. Now do GramSchmidt.




#7
Nov1308, 11:57 PM

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#8
Nov1408, 12:13 AM

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P: 21,063

a + b + c = 0 If you solve for a, you get
To make things even more explicit,
[tex] \left[ \begin{array}{ c } a \\ b \\ c \end{array} \right] = b\left[ \begin{array}{ c } 1 \\ 1 \\ 0 \end{array} \right] + c\left[ \begin{array}{ c } 1 \\ 0 \\ 1 \end{array} \right][/tex] Voila, there's your basis. 



#9
Nov1408, 04:17 AM

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PF Gold
P: 38,898





#10
Jul2710, 12:09 AM

P: 123

I'm currently doing the same problem so I figured I'd revive this old thread.
Once I have my two vectors, u1=<1,1,0> and u2=<1,0,1> I then go through the GramSchmidt process to find the normalized basis? 



#11
Jul2710, 01:08 AM

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P: 21,063

If you already have a basis, you don't need GramSchmidt to find a normalized basis.
Do you know what the term "normalized" means? 



#13
Jul2710, 02:15 PM

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P: 21,063

OK, in that case you need to use GS.



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