Solving Orthogonal Matrix Homework w/ Symmetric Matrix

[beetle2]

Homework Statement

Given the symmetric Matrix

1 2
2 5

find an orthogonal matrix P such that C=BAB^t

Homework Equations

The Attempt at a Solution

I found the eigenvalues to be 3-(2\sqrt{2}) and 3+(2\sqrt{2})

giving eigenvectors of
<br /> [1,1-\sqrt{2}] and [1,1+\sqrt{2}]

As the dot product of these vectors is 0 they are orthogonal.

do I just normalise each vector and use them as the column vectors of P?

 

[aPhilosopher]

It's going to be very difficult to make a statement about C=BAB^t in general. Knowing a symmetric matrix P, associated with an unlabeled matrix does very little to help.

 

[beetle2]

Sorry the matrix is

A =

1 2
2 5

find an orthogonal matrix P such that C=PAP^t where C is diagonal

regards

 

[HallsofIvy]

Then find the eigenvectors of A. P will be an orthogonal matrix with the eigenvectors of A as rows.

 

[beetle2]

So I multiply P =
<br /> \[ \left( \begin{array}{cc}<br /> 1 &amp; 1-\sqrt{2} \\<br /> 1 &amp; 1+\sqrt{2} \\<br /> \end{array} \right)\] <br />
by A =
<br /> \[ \left( \begin{array}{cc}<br /> 1 &amp; 2 \\<br /> 2 &amp; 5 \\<br /> \end{array} \right)\] <br />
and P^T =
<br /> \[ \left( \begin{array}{cc}<br /> 1 &amp; 1 \\<br /> 1-\sqrt{2} &amp; 1+\sqrt{2} \\<br /> \end{array} \right)\] <br />

which gives C =

<br /> \[ \left( \begin{array}{cc}<br /> 20-14\sqrt{2} &amp; 0 \\<br /> 0 &amp; 14\sqrt{2}+20 \\<br /> \end{array} \right)\] <br />


Is this right? I know that C is diagonal but isn't it supposed to have the eigenvalues on the main diagonal?
regards