## Deflation of the largest eigenvalue

$$\begin{pmatrix} 1&2&1&2\\ 2&1&2&1\\ 1&2&1&2\\ 2&1&2&1 \end{pmatrix}$$
The matrix $$\mathbf A$$ has eigenvalues 6, 0 (multiplicity two) nd $$-2$$. The eigenvalue $$\lambda_1=6$$ corresponds to eigenvector $$\mathbf x_1=(1{,}1,1{,}1)^T$$, we normalize it to $$\mathbf z_1=\frac12(1{,}1,1{,}1)^T$$.
By the deflation we get the matrix $$\mathbf A'= \mathbf A - \lambda_1 \mathbf z_1 \mathbf z_1^T = \begin{pmatrix} 1&2&1&2\\ 2&1&2&1\\ 1&2&1&2\\ 2&1&2&1 \end{pmatrix} - 6\cdot \frac12 \begin{pmatrix} 1\\ 1\\ 1\\ 1 \end{pmatrix} \cdot \frac12 \begin{pmatrix} 1&1&1&1 \end{pmatrix} =\frac12 \begin{pmatrix} -1&1&-1&1\\ 1&-1&1&-1\\ -1&1&-1&1\\ 1&-1&1&-1 \end{pmatrix}$$
The matrix $$\mathbf A'$$ has eigenvalues 0 of multiplicity 3 and $$-2$$ and the same eigenvectors as the matrix $$\mathbf A$$.