How do you orthogonally diagonalize the following matrix:

$\displaystyle

A=\begin{pmatrix}1&0&0\\0&3&-2\\0&-2&3\end{pmatrix}

$

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- Apr 20th 2009, 07:07 AMantmanOrthogonally diagonalize the matrix
How do you orthogonally diagonalize the following matrix:

$\displaystyle

A=\begin{pmatrix}1&0&0\\0&3&-2\\0&-2&3\end{pmatrix}

$ - Apr 20th 2009, 07:53 AMTwighi
hi

You need to find the eigenvalues of the matrix, and find the corresponding eigenvectors. Do you know how to do this?

$\displaystyle A = PDP^{T} $ , since A is orthogonally diagonizable, P will be an orthogonal matrix, hence $\displaystyle P^{-1}=P^{T} $ .

An easy way to find the eigenvalues of this matrix would be a cofactor expansion for example, or you could use the fact that it is block triangular.

Itīs quite a long process to describe if you have no idea whatsoever on how to find eigenvalues or eigenvectors, and I would in that case suggest you start there. - Apr 21st 2009, 12:16 PMantman
I think the eigenvalues are 5, 1 and 1 or just 5 and 1? and the eigenvectors are <0 -1 1>, <0 1 1> and <1 0 0>?

- Apr 21st 2009, 12:55 PMTwighi
hi

Yes, the eigenvalus are $\displaystyle \lambda_{1} = 5 \; \lambda_{2} = 1 \; \lambda_{3} = 1 $ so the multiplicity of eigenvalue 1 is two. This means that the dimension of the nullspace of the matrix $\displaystyle (A- 1 \cdot I) $ has dimension 2.

The eigenvectors you wrote are correct.

There is a theorem that says that eigenvectors from different eigenspaces to a symmetric matrix are orthogonal, so you only need to verify that the two vectors corresponding to eigenvalue 1 are orthogonal to each other.

Which they are here.

Now you normalize your eigenvectors and put them in a matrix P, and the eigenvalues you put in a diagonal matrix D, with the eigenvalue corresponding to the first column vector in the first column of D etc. Order matters here!

And you are done! - Apr 21st 2009, 01:08 PMantman
My P matrix is a little ugly but I'm hoping it is correct.

$\displaystyle

\begin{pmatrix}0&1&0\\-\sqrt2/2&0&\sqrt2/2\\\sqrt2/2&0&\sqrt2/2\end{pmatrix}

$ - Apr 21st 2009, 01:28 PMantman
and my final would be the eigenvectors on the diagonal like $\displaystyle \begin{pmatrix}5&0&0\\0&1&0\\0&0&1\end{pmatrix}$?

- Apr 22nd 2009, 02:39 AMTwighi
hi

Yes, your matrix P is correct, and so is your matrix D.

$\displaystyle A = PDP^{T} $

I assumed you meant eigenVALUES on the diagonal, not eigenvectors (Happy)

good job