# Orthogonal MAtrix

1. Oct 27, 2009

### beetle2

1. The problem statement, all variables and given/known data
Given the symmetric Matrix

1 2
2 5

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

2. Relevant equations

3. The attempt at a solution

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

giving eigenvectors of
$[1,1-\sqrt{2}]$ 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 27, 2009

### 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.

3. Oct 27, 2009

### 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

4. Oct 28, 2009

### HallsofIvy

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

5. Oct 28, 2009

### beetle2

So I multiply P =
$$\left( \begin{array}{cc} 1 & 1-\sqrt{2} \\ 1 & 1+\sqrt{2} \\ \end{array} \right)$$
by A =
$$\left( \begin{array}{cc} 1 & 2 \\ 2 & 5 \\ \end{array} \right)$$
and PT =
$$\left( \begin{array}{cc} 1 & 1 \\ 1-\sqrt{2} & 1+\sqrt{2} \\ \end{array} \right)$$

which gives C =

$$\left( \begin{array}{cc} 20-14\sqrt{2} & 0 \\ 0 & 14\sqrt{2}+20 \\ \end{array} \right)$$

Is this right? I know that C is diagonal but isnt it supposed to have the eigenvalues on the main diagonal?
regards