## Diagonalization of a symmetric matrix

Diagonalize the following symmetric matrix $$\mathbf A=\begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end{pmatrix}$$ in the form $$\mathbf A=\mathbf Q\mathbf\Lambda \mathbf Q^T$$.
• #### Solution

The procedure is the same as for normal diagonalization of matrices (a real symmetric matrix has real eigenvalues and is diagonalizable), only the basis composed of eigenvectors must be orthonormal.

For the matrix $$\mathbf A$$ we get eigenvalues $$\lambda_{1{,}2}=0$$ a $$\lambda_3=3$$.

The eigenvalue 0 corresponds e.g. to eigenvectors $$\mathbf x_1=(1{,}0,-1)^T$$ a $$\mathbf x_2=(0{,}1,-1)^T$$, the eigenvalue 3 corresponds to an eigenvector $$\mathbf x_3=(1{,}1,1)^T$$.

Observe that the vector $$\mathbf x_3$$ is perpendicular to $$\mathbf x_1,\mathbf x_2$$, so it suffices to normalize it. Vectors $$\mathbf x_1,\mathbf x_2$$ shall be transformed by Gram-Schmidt ortogonalization:

$$\begin{eqnarray*} \mathbf x_1&=&(1{,}0,-1)^T \\ \mathbf z_1&=&\frac{1}{\sqrt{2}}(1{,}0,-1)^T\\ \mathbf x_2&=&(0{,}1,-1)^T\\ \mathbf y_2&=&\mathbf x_2-\langle \mathbf x_2|\mathbf z_1\rangle \mathbf z_1=\dots = \frac{1}{1}(-1{,}2,-1)^T\\ \mathbf z_2&=&\frac{1}{\sqrt{6}}(-1{,}2,-1)^T\\ \mathbf x_3&=&(1{,}1,1)^T\\ \mathbf z_3&=&\frac{1}{\sqrt{3}}(1{,}1,1)^T \end{eqnarray*}$$
$$\mathbf \Lambda=\begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&3 \end{pmatrix}$$, $$\mathbf Q=\frac{1}{\sqrt{6}} \begin{pmatrix} \sqrt{3}&-1&\sqrt{2}\\ 0&2&\sqrt{2}\\ -\sqrt{3}&-1&\sqrt{2} \end{pmatrix}$$