= Solution
Using the monic convention for the <characteristic polynomial>, expand the <determinant> along the last column:
$$
\boxed{\chi_A(t)=\det(tI-A)=(t-2)\det\begin{pmatrix}t&-1\\1&t-2\end{pmatrix}=(t-2)(t-1)^2}.
$$
For the <eigenvalue> $1$, the equations $(A-I)v=0$ give $v_2=v_1$ and $v_3=0$. Thus every corresponding nonzero <eigenvector> is a nonzero multiple of $(1,1,0)^T$. For the <eigenvalue> $2$, the first two equations give $v_2=2v_1$ and $v_1=0$, so every corresponding nonzero <eigenvector> is a nonzero multiple of $(0,0,1)^T$. Therefore
$$
\boxed{E_1=\operatorname{span}\{(1,1,0)^T\},\qquad E_2=\operatorname{span}\{(0,0,1)^T\}}.
$$
These <eigenspaces> together have dimension $2$, so they cannot provide an <eigenvector> <basis> of $\mathbb R^3$. \b[The <matrix> is not diagonalizable]; the <eigenvalue> $1$ has algebraic multiplicity $2$ but geometric multiplicity $1$.
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