Orthogonal diagonalization of a symmetric matrix

Eigenvalues. $\det(A - \lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-1)(\lambda-3)$, so the eigenvalues are $1$ and $3$.

Eigenvectors. For $\lambda = 1$: $v_1 = (1,-1)$. For $\lambda = 3$: $v_2 = (1,1)$. These are already orthogonal.

Normalization. $\|v_1\| = \sqrt2$, $\|v_2\| = \sqrt2$, so
$$u_1 = \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \qquad u_2 = \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$

Orthogonal matrix $Q$. Let
$$Q = \frac{1}{\sqrt2}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}.$$
Then $Q^T Q = I$ and
$$Q^T A Q = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}.$$

Conclusion. $A = Q \operatorname{diag}(1,3) Q^T$ with $Q$ orthogonal.