Orthogonal Diagonalization of $~\begin{bmatrix} 9&-1&-2\\ -1&9&-2\\ -2&-2&6\end{bmatrix}~$ with repeated eigenvalue

Question

Consider a matrix $$ A=\begin{bmatrix} 9&-1&-2\\ -1&9&-2\\ -2&-2&6\end{bmatrix}. $$ We now find the orthogonal matrix $P$ such that $P^TAP$ is diagonal.

My attempt:

Eigenvalues are $4$ and $10$ (with multiplicity $2$). Eigenvectors corresponding to $4$ are $~c \begin{bmatrix} 1\\1\\2\end{bmatrix},~c \neq 0~$ and corresponding to $10~$ are $~c \begin{bmatrix} 1\\0\\-\dfrac{1}{2}\end{bmatrix}+ d\begin{bmatrix} 0\\1\\-\dfrac{1}{2}\end{bmatrix},~c,d \neq 0.$ Now from here how can I get an orthogonal matrix so that $~P^TAP~$ is diagonal? If the question is about invertible matrix, the we choose $~P=\begin{bmatrix} 1&1&0\\0&1&1\\ 2&-\frac{1}{2}& -\frac{1}{2}\end{bmatrix}.$ But how can I get an orthogonal matrix from here?

Please help me to solve this.

Answer 1

As it is $3\times 3$ one can use this trick.

Take a $4$-eigenvector, $a:=(1,2,2)$.

Take any $10$-eigenvector, $b:=(2,0,-1)$ say.

These are bound to be orthogonal to each other.

They are also orthogonal to $a\times b=(-1,5,-2)$, and that is bound to be your third vector.

Now all you need to do is normalise these.

Answer 2

Yes, the eigenvectors corresponding to the eigenvalue $10$ are the non-null vectors of the form$$\begin{bmatrix}c\\d\\-\frac{c+d}2\end{bmatrix}.$$Now, take two of these vectors which are orthogonal, such as$$\begin{bmatrix}-2\\0\\1\end{bmatrix}\quad\text{and}\quad\begin{bmatrix}-1\\5\\-2\end{bmatrix}.$$Now, since indeed$$\begin{bmatrix}1\\1\\2\end{bmatrix}$$is an eigenvector correponding to the eigenvalue $4$, take$$P=\begin{bmatrix}-\frac2{\sqrt5}&-\frac1{\sqrt{30}}&\frac1{\sqrt6}\\0&\frac5{\sqrt{30}}&\frac1{\sqrt6}\\\frac1{\sqrt5}&-\frac2{\sqrt{30}}&\frac2{\sqrt6}\end{bmatrix}:$$the columns of $P$ are the $3$ vectors from above after normalization (that is, after dividing each of them by its norm).