Question 5.2.3: Let A = [ 0 1 1 1 0 1 1 1 0] and B = [ -1 1 0 0 -1 1 0 0 2 ]...

To find the eigenvalues of A, we solve the characteristic equation
det(A − λI) = det \begin{vmatrix} −λ& 1 &1\\1 &−λ& 1\\1& 1& −λ \end{vmatrix} = −(λ + 1)² (λ − 2) = 0
Thus, the eigenvalues of A are \lambda _{1} = −1, with algebraic multiplicity 2, and \lambda _{2} = 2, with algebraic multiplicity 1. To find the eigenvectors, we find the null space of A − λI for each eigenvalue. For \lambda _{1} = −1 we reduce the matrix

Hence,
N(A + I) = span \left\{\begin{bmatrix} -1 \\ 1 \\0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\1 \end{bmatrix} \right\}

In a similar manner we find
N(A − 2I) = span \left\{\begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} \right\}

Since the three vectors \begin{bmatrix} -1 \\ 1 \\0 \end{bmatrix} , \begin{bmatrix} -1 \\ 0 \\1 \end{bmatrix}  and \begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} are linearly independent, by Theorem 2 the matrix A is diagonalizable.
Using the same approach, we find that B has the same characteristic polynomial and hence the same eigenvalues. However, in this case

N(B + I) = span \left\{\begin{bmatrix} 1 \\ 0 \\0 \end{bmatrix} \right\}      and     N(B −2I) =span \left\{\begin{bmatrix} 1 \\ 3 \\9 \end{bmatrix} \right\}

Since B does not have three linearly independent eigenvectors, by Theorem 2, B is not diagonalizable