linear algebra - How to find orthogonal eigenvectors if some of the eigenvalues are the same?

I have an example:
$$A=\begin{pmatrix} 2 & 2 & 4 \\ 2 & 5 & 8 \\ 4 & 8 & 17 \end{pmatrix}$$
The eigenvalue I found is $\lambda_1=\lambda_2=1$ and $\lambda_3=22$.
For $\lambda=1$,
$$\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} -2\\ 1 \\ 0 \end{pmatrix}y+\begin{pmatrix} -4\\ 0 \\ 1 \end{pmatrix}z$$
For $\lambda=22$,
$$\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} 1/4\\ 1/2 \\ 1 \end{pmatrix}z$$

However, those eigenvectors I found are not orthogonal to each other. The goal is to find an orthogonal matrix P and diagonal matrix Q so that $A=PQP^T$.

Answer

One thing we know is that eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal. So, if we find eigenvectors $v_1,v_2,v_3$ for $\lambda_1< \lambda_2< \lambda_3$ we are done. On the other hand, we have eigenvalues $\lambda_1=\lambda_2=1$ and $\lambda_3=22$, so that there are not $3$ distinct eigenvalues and the situation becomes somewhat more complicated.

Suppose we found $v_1,v_2\in E(A,\lambda_1)$ which are linearly independent (and hence a basis for the Eigenspace). We know that $v_1\perp v_3$ and $v_2\perp v_3$. This means $\langle v_1,v_3\rangle=\langle v_2,v_3\rangle=0$. By bilinearity of the inner product, we get that $\langle av_1+bv_2,v_3\rangle =0$ for all $a,b\in \mathbb{R}$. The upshot is that the entire eigenspace $E(A,\lambda_1)$ is orthogonal to $v_3$. So, we are free to choose any basis of eigenvectors for $E(A,\lambda_1)$ and proceed from there. Well, just apply Gram-Schmidt to $v_1,v_2$. Define
$$u_1=\frac{v_1}{\lVert v_1\rVert}$$
$$u_2=\frac{v_2-\langle v_2, u_1\rangle u_1}{\lVert v_2-\langle v_2, u_1\rangle u_1\rVert}.$$
A quick check shows that these two vectors form an orthonormal basis for $E(A,\lambda_1)$. Then, if we take any nonzero $v_3\in E(A,\lambda_3)$ and set
$$u_3=\frac{v_3}{\lVert v_3\rVert}$$
we can see that $(u_1,u_2,u_3)$is an orthonormal eigenbasis of $\mathbb{R}^3\cong E(\lambda_1,A)\oplus E(\lambda_3,A)$ with respect to $A$. You've already found the vectors $v_1,v_2,v_3$. Once you compute $u_1,u_2,u_3$, the matrix $P=[u_1,u_2,u_3]$ is orthogonal and

$$A=P^T \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&22 \end{bmatrix} P.$$