math
posted by john .
Find an orthonormal basis for the subspace of R^3 consisting of all vectors(a, b, c) such that a+b+c = 0.
The subspace is twodimensional, so you can solve the problem by finding one vector that satisfies the equation and then by constructing another solution which is orthogonal to it.
E.g. if you take the first basis to be (1,1,0) up to normalization, then the second basis vector (a,b,c) must have inner product zero with the first basis vector. Therefore:
a  b = 0
We can take, e.g. a = 1 and b = 1.
a + b + c = 0, therefore c = 2
You only have to normalize these two vectors.
In general linear algebra problems you should avoid this way of solving problems, i.e. solving equations to find orthonormal vectors. In this case it was easy, because we only had to find two vectors.
So, let's do this problem again using the sandard way
Let's start by writing down two arbitrary independent vectors, e.g.:
e1 = (1,1,0)
e2 = (1,0,1).
Then you use the GramSchmidt method: Take the first normalized basis vector to be:
f1 = e1/e1 = 1/sqrt[2] (1,1,0)
Subtract from e2 the projection of e2 on f1 to obtain a vector that is orthogonal to f1:
e2' = e2  <e2,f1>f1.
The fact that e2' is orthogonal to f1 is easy to check:
<e2',f1> = <(e2  <e2,f1>f1),f1> =
<e2,f1 >  <e2,f1><f1,f1>
Because f1 is normalized to 1,
<f1,f1> (which always equals the square of the norm) equals 1 and we see that <e2',f1> = 0
We can thus take the second orthonormal basis vector to be:
f2 = e2'/e2'
e2' = e2  <e2,f1>f1 =
(1,0,1)  1/2 (1,1,0) =
(1/2,1/2,1)
f2 = sqrt[1/6] (1,1,2)
So, no equation solving is necessary to find the orthonormal set from a set of linear independent vectors.
If you had to orthonormalize a larger set of independent vectors, then the next step would have been:
e3' = e3  <e3,f1>f1  <e3,f2>f2
f3 = e3'/e3'
e4'= e3  <e3,f1>f1  <e3,f2>f2<e3,f3>f3
f4 = e4'/e4'
etc.
Respond to this Question
Similar Questions

math
There is one step in a proof that I don't understand. Could someone please explain? 
Algebra
How would you prove this theorem: The column space of an m x n matrix A is a subspace of R^m by using this definition: A subspace of a vector space V is a subset H of V that has three properties: a) the zero vector of V is in H. b) … 
linear algebra
which of the following sets of vectors span R^3? 
math
A trigonmetric polynomial of order n is t(x) = c0 + c1 * cos x + c2 * cos 2x + ... + cn * cos nx + d1 * sin x + d2 * sin 2x + ... + dn * sin nx The output vector space of such a function has the vector basis: { 1, cos x, cos 2x, ..., … 
math
Find the least squares approximation of x over the interval [0,1] by a polynomial of the form a + b*e^x  The polynomial produces an output space with two linearly independent … 
linear algebra
what is the basis of a subspace or R3 defined by the equation 2x1+3x2+x3=0 I know the basis is vectors <3,2,0> and <1,0,2> but how do you get that? 
Linear Algebra
1/ Prove that the set V=R+ ( the set of all positive real numbers) is a vector space with the following nonstandard operations: for any x,y belong to R+ & for any scalar c belong to R: x O+ ( +signal into circle) y=x.y (definition … 
Math
Went ahead and did the HW the teach recommended but she did not post the answers and I would like to see if im on the right track. Problem 1: Are the vectors (2,−1,−3), (3, 0,−2), (1, 1,−4) linearly independent? 
Math Elementary Linear Algebra
determine whether or not the given set forms a basis for the indicated subspace: #1 {1,1,0), (1,1,1)} for the subspace of R^3 of all (x,y,z) such that y= x+z #2 {[1,2,1,3), (0,0,0,0)} for the subspace of R^4 of all vectors of the … 
Math
Let V and W be vector spaces and assume that T be a linear mapping from V to W. (a) Show that ker(T) is a subspace of V . (b) Show that range(T) is a subspace of W.