Diagonalizable by orthonormal matrix Given the matrix
$$A = \begin{bmatrix}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2 \end{bmatrix}$$
Explain why $A$ can be diagonalized by an orthonormal matrix and find an orthonormal matrix $Q$ and a diagonal $D$ such that $A = QDQ^T$.
My approach : Seems like a problem that goes around the eigenvalue-eigenmatrix thing. So, $x_A(λ)=\det(λI-A) = -(λ-4)(λ-1)^2$ is the characteristic polynomial. We get that the eigenvalues are : $λ_1 = 4$, $λ_2=λ_3=1$. It's easy to find the eigenvectors then, that are : $v_1 = (1,1,1)^T$, $v_2=(1,0,-1)^T$, $v_3=(-1,1,0)^T$. We can proceed after that to the standard diagonalization form, with the diagonal matrix formed by the eigenvalues and the orthonormal matrices formed by the eigenvectors.
Is my approach correct? HOW do I explain first of all why can $A$ be diagonalized by an orthonormal matrix ? After that, the question that follows is answered by my solution ? I'd really appreciate some help over this, it's an old exam question that I am looking over for my end-terms.
 A: Your approach is completely correct, and you can justify it by remembering:
1) A symmetric matrix is always diagonalizable and all its eigenvalues are real;
2) For a symmetric matrix, eigenvectors belonging to different eigenvalues are orthogonal
The last property is exactly what allows you to orthogonally diagonalize your matrix: observe that eigenvectors belonging to the different eigenvalues of the matrix are not only linearly independent but in fact orthogonal. Now you only need to apply Gram-Schmidt within each eigenspace separatedly, and that's all.
A: The reason $A$ can be diagonalized by an orthogonal matrix is simply that it is symmetric.
There is one small problem with your solution : if you want the matrix formed by the eigenvectors to be orthogonal, you need to take orthonormal vectors. You can apply the Gram-Schmidt process to make the eigenvectors orthonormal; notice that since $v_1$ is orthogonal to $v_2$ and $v_3$ (which is expected since they are associated to different eigenvalues), you will only need to change $v_3$ and divide the eigenvectors by their norms.
