Find orthogonal matrix $P$ such that $P^tAP = D$ for $A$ is a normal operator For 
$$A = \;\;\;
        \begin{pmatrix}
        1 & 2 & 3 \\
        2 & 3 & 4 \\
        3 & 4 & 5 \\
        \end{pmatrix}
$$
there is a real orthogonal matrix $P$ such that $P^tAP = D$ is diagonal. Find such a diagonal matrix $D$.

I know as $A$ is a normal operator, there is a real diagonal matrix corresponding to $A$. Firstly, I tried to find eigenvalue and eigen vectors to find orthogonal matrix $P$, however, such a $P$ is not orthogonal.
Can anybody help? Thanks!

Hints:


*

*Find the eigenvalues

*Find the eigenvectors

*Normalize the eigenvectors and take $P$ to be these normalized eigenvectors, which gives $P^{-1} = P^t$


This is possible since this is a real, symmetric matrix that has three distinct eigenvalues. If that were not the case, this algorithm would need to be modified slightly and this process is not always possible.
For this problem, we have three real and distinct eigenvalues, so we can straight off write the diagonal matrix as:
$$J =  \begin{bmatrix}
        0 & 0 & 0 \\
        0 & \dfrac{1}{2}\left(9-\sqrt{105}\right) & 0 \\
        0 & 0 & \dfrac{1}{2}\left(9+\sqrt{105}\right) \\
        \end{bmatrix}
$$
