# Finding an orthonormal basis

by DWill
Tags: basis, orthonormal
 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 Gram-Schmidt 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!
 Sci Advisor HW Helper Thanks P: 25,174 If you know Gram-Schmidt, 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 Gram-Schmidt with? (1,-1,0) sounds like a good start. Give me another one.
 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?
HW Helper
Thanks
P: 25,174

## 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.
 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?
 Sci Advisor HW Helper Thanks P: 25,174 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 Gram-Schmidt.
Mentor
P: 21,063
 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 Gram-Schmidt process to make them orthonormal.
If you apply Gram-Schmidt to these vectors, you'll just get the same ones. They are already orthogonal and have length 1.
Mentor
P: 21,063
 Quote by DWill 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?
I'm not sure it's clear to you where Dick got the vectors he did. Start with your equation:
a + b + c = 0

If you solve for a, you get
a = -b - c, where b and c are arbitrary
b =  b
c =      c
That last two equations are obviously true.
To make things even more explicit,
a = -1b - 1c
b =  1b + 0c
c =  0b  +1 c
or,
$$\left[ \begin{array}{ c } a \\ b \\ c \end{array} \right] = b\left[ \begin{array}{ c } -1 \\ 1 \\ 0 \end{array} \right] + c\left[ \begin{array}{ c } -1 \\ 0 \\ 1 \end{array} \right]$$
Math
Emeritus
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PF Gold
P: 38,898
 Quote by DWill 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?
You need to review basic definitions. Those two vectors CAN'T be a basis for R^3 and can't span R^3 because that requires THREE vectors. You must have misread the original problem. As Dick said, the set of all (a, b, c) such that a+ b+ c= 0 is a two dimensional subspace of R^3. NO set of such vectors can be a basis for R^3. (1, -1, 0) and (1, 0, -1) form a basis for that two dimensional subspace.
 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 Gram-Schmidt process to find the normalized basis?
 Mentor P: 21,063 If you already have a basis, you don't need Gram-Schmidt to find a normalized basis. Do you know what the term "normalized" means?
P: 123
 Quote by Mark44 If you already have a basis, you don't need Gram-Schmidt to find a normalized basis. Do you know what the term "normalized" means?
Sorry, I meant a orthonormal basis.
 Mentor P: 21,063 OK, in that case you need to use G-S.

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