# Orthogonally diagonalise a 3x3 matrix

• Feb 9th 2013, 10:13 AM
marky82
Orthogonally diagonalise a 3x3 matrix
I've been at this for a while and I'm now stumped.

The matrix, A is

5 -4 0
-4 3 -4
0 - 4 1

Here's what I've got so far.

eigenvalues are λ = 9, λ = -3, λ = 3

corresponding eigen vectors are (2k, -2k, k), (k/2, k, k), (-k, k/2, k)

{(2, -2, 1)} is a basis for s(9)
{(1,2,2)} is a basis for s(-3)
{(-2,-1,2)} is a basis for s(3)

So that gives the set

E = {(2,-2,1),(1,2,2),(-2,-1,2)} which is an eigenvector basis of A (not sure if this is even needed as I think I need an orthonormal basis)

I think I need to find an orthonormal eigenvector basis next, but I can't understand how to do that (if that is actually next). As that should give me the transition matrix.

So looking for someone to hold my hand to the end of this process please :)
• Feb 9th 2013, 10:31 AM
ILikeSerena
Re: Orthogonally diagonalise a 3x3 matrix
Hi marky82! :)

An orthonormal basis requires 2 things: orthogonal and normal.
Orthogonal means that he vectors are orthogonal.
Normal means that they have length 1.

And what is their length?
• Feb 9th 2013, 10:49 AM
marky82
Re: Orthogonally diagonalise a 3x3 matrix
Quote:

Originally Posted by ILikeSerena
Hi marky82! :)

An orthonormal basis requires 2 things: orthogonal and normal.
Orthogonal means that he vectors are orthogonal.
Normal means that they have length 1.

And what is their length?

Hello again :)

the eigenvectors are othogonal as they correspond to the distinct eigenvalues of the symmetric matrix above.

I have no idea on how to find the length so not sure what is next :P
• Feb 9th 2013, 11:05 AM
ILikeSerena
Re: Orthogonally diagonalise a 3x3 matrix
Two vectors are orthogonal if they are perpendicular.
This is equivalent to their dot product being zero.

What are the dot products of each pair of vectors in the basis?

The (regular) length of a vector is $\sqrt{x^2+y^2+z^2}$
What are the lengths of the vectors in your basis?

To normalize a vector, you divide each of its coordinates by its length...
• Feb 10th 2013, 02:52 AM
marky82
Re: Orthogonally diagonalise a 3x3 matrix
Quote:

Originally Posted by ILikeSerena
Two vectors are orthogonal if they are perpendicular.
This is equivalent to their dot product being zero.

What are the dot products of each pair of vectors in the basis?

The (regular) length of a vector is $\sqrt{x^2+y^2+z^2}$
What are the lengths of the vectors in your basis?

To normalize a vector, you divide each of its coordinates by its length...

Sorry for the late reply, I decided to give my brain a break and have another go at it today, to see if I can finish it :)