= Solution
For $A$, the <triangular matrix> formula gives the <characteristic polynomial>
$$
\det(tI-A)=(t-1)(t-3)(t-2).
$$
The three <eigenvalues> are distinct, so part (d) already proves the existence of an <eigenvector> <basis>. Explicit choices, ordered by <eigenvalues> $1,3,2$, are
$$
\boxed{(1,0,0)^T,\quad(1,1,0)^T,\quad(2,-1,1)^T.}
$$
Their column <matrix> has <determinant> one, and direct multiplication gives respectively one, three and two times the indicated <vector>.
For $B$, expansion along the last row gives
$$
\det(tI-B)=(t-3)((t-3)^2-1)=(t-3)(t-4)(t-2).
$$
Again the three <eigenvalues> are distinct. An explicit <eigenvector> <basis>, ordered by <eigenvalues> $4,2,3$, is
$$
\boxed{(1,1,0)^T,\quad(1,-1,0)^T,\quad(0,-1,1)^T.}
$$
Its column <matrix> has <determinant> $-2\ne0$. In particular, $B$'s diagonal entries are not its <eigenvalues>: it is not a <triangular matrix>.
Finally $C=I_3$, so every nonzero <vector> is an <eigenvector> with <eigenvalue> one. Choose \b[the <standard basis> $e_1,e_2,e_3$]. These are <linearly independent> even though their <eigenvalues> coincide. Distinct <eigenvalues> are sufficient for an <eigenvector> <basis>, and are not necessary.
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