Orthogonal diagonalization of a symmetric matrix

Eigenvalues. $\det(A-\lambda I)=(3-\lambda)^2-1=\lambda^2-6\lambda+8=(\lambda-2)(\lambda-4)$. So $\lambda_1=2$ and $\lambda_2=4$.

Eigenvectors.
- For $\lambda=2$: $v_1=(1,-1)$.
- For $\lambda=4$: $v_2=(1,1)$.

These are already orthogonal.

Normalization. $u_1=(1,-1)/\sqrt2$ and $u_2=(1,1)/\sqrt2$.

Orthogonal matrix $Q$.
$$Q = \frac{1}{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}.$$
Then $Q^TQ=I$ and
$$Q^TAQ = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}.$$

Conclusion. $A=Q\textrm{diag}(2,4)Q^T$ with $Q$ orthogonal.