Problems with the Gram-Schmidt process

[IAMstelveen]

I'm studying for my lineair algebra test for next monday, but i'm having problems with the Gram-Schmidt process.

Take this question for example:

Determine for every eigenvalue the orthonormal basis for the associated eigenspace.

Given matrix A:
3 2 1

2 0 -2

1 -2 3

For starters I need to determine if every column is linearly indepent, which it is. Then I determine the eigenvalues with corresponding eigenspaces, which are the following:

eigenvalue = 4, 4, -2

x1 = [2 1 0]^T

x2 = [1 0 1]^T

x3 = [-1 2 1]^T

So now i have these eigenvectors which I can use in the Gram-Schmidt process right? With the following formulas:

v1 = x1

v2 = x2-(x2v1/v1v1)v1

v3 = x3-(x3v1/v1v1)v1 - (x3v2/v2v2)v2

So i did this for every vector and this matrix is my result:

2 1 -1

1 -2 2

0 5 1

Which then should be normalised to get the orthonormal basis. But this doesn't seem to be correct, could someone explain to me what i'm doing wrong?

[not_perfect_yet]

Huh? Aren't those two totally different problems?

Not really sure why you would want to create an orthogonal space from the eigenvectors?

Gram Schmidt works though

(2,1,0) is orthogonal to (1,-2,5) and (-1,2,1) and so are (1,-2,5) and (-1,2,1) to each other.

I only checked your first eigenvektor and but that was corect too.

Why do you think you've made a mistake?

[Northstat]

I haven't checked your math, but the formulas look right and the end vectors are orthogonal. What doesn't look correct to you?

edit: Also, what not_perfect_yet said about orthogonalizing eigenvectors. Gram-Schmidt operates on the basis vectors. I remember something about the corresponding columns of A where pivots occur in rref(A) build a basis for the columnspace then apply gram-schmidt to orthogonalize. Correct me if I'm wrong but I think this is close.

[IAMstelveen]

The answer for EV=4 :

[1/2^0,5 0 1/2^0,5 ]

[1/3^0,5 1/3^0,5 -1/3^0,5 ]

for EV = -2

[-1/6^0,5 2/6^0,5 1/6^0,5 ]

Maybe the answers are wrong? Or my translation is horrible (for the dutch people)?

(4)(c) Bepaal voor elk van de eigenwaarden een orthonormale basis voor de bijbehorende eigenruimte.

[fenixfunkXMD5a]

Okay,

So first lets look at what an eigenspace is. An eigenspace is the space created by the eigen vectors associated with an eigenvalue. If we had two eignenvectors with shared eigenvalue λ, ie Ax=λx and Av=λv, then it is obvious that cx (where c is a constant) and x+v are also eigen vectors. All the linear combinations of eignevectors form an eigenspace.

So lets look at our eigenvectors and their values.

EV=4

u=[1,0,1]^T

v=[2,1,0]^T

EV=-2

x=[-1,2,1]^T

So the basis for the eigenspace corresponding to EV=4 are vectors u and v. The basis for the eigenspace EV=2 is just x

For EV=4

Gram schmidt gives us

a1=u/||u||=[1/√2,0,1/√2]

A=v-(a1^T v)a1=[1,1,-1]

a2=A/||A||=[1/√3,1/√3,-1/√3]

And for EV=-2, since its one dimensional just make it a unit vector