= Solution
Expanding the <characteristic polynomial> gives
$$
\det(\lambda I-A)=\lambda^3-2\lambda^2-\lambda+2
=(\lambda+1)(\lambda-1)(\lambda-2).
$$
Solving the three nullspace systems gives \b[the <eigenvalues> and <eigenspaces>]
$$
\boxed{E_{-1}=\operatorname{span}\{(1,1,1)^T\},\quad
E_1=\operatorname{span}\{(1,1,0)^T\},\quad
E_2=\operatorname{span}\{(0,1,-1)^T\}.}
$$
For example, at $\lambda=1$ the first two equations force $z=0$ and $x=y$; at $\lambda=2$ they force $x=0$ and $y=-z$; at $\lambda=-1$ they force $x=y=z$.
Putting these <eigenvectors> in columns proves their <linear independence>:
$$
P=\begin{pmatrix}1&1&0\\1&1&1\\1&0&-1\end{pmatrix},\qquad\det P=1,\qquad
P^{-1}AP=\operatorname{diag}(-1,1,2).
$$
Hence \b[$A$ is a <diagonalizable matrix> over the reals].
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