For , the triangular matrix formula gives the characteristic polynomial
The three eigenvalues are distinct, so part (d) already proves the existence of an eigenvector basis. Explicit choices, ordered by eigenvalues , are
Their column matrix has determinant one, and direct multiplication gives respectively one, three and two times the indicated vector.
For , expansion along the last row gives
Again the three eigenvalues are distinct. An explicit eigenvector basis, ordered by eigenvalues , is
Its column matrix has determinant . In particular, 's diagonal entries are not its eigenvalues: it is not a triangular matrix.
Finally , so every nonzero vector is an eigenvector with eigenvalue one. Choose the standard basis . 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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