Finding an orthonormal basis

[DWill]

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!

[Dick]

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.

[DWill]

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?

[Dick]

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

[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?

[Dick]

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.

[Mark44]

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.

[Mark44]

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]
Voila, there's your basis.

[HallsofIvy]

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.

[jinksys]

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?

[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?

[jinksys]

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.

[Mark44]

OK, in that case you need to use G-S.