Discussion
Let \(A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 2 & 0 \\ 1 & 1 & 4 \end{bmatrix}\). Then, the eigenvalues of the matrix \(A^{2} - 3A + 2I\) are :
Answer: A
The eigenvalues are 0, 0, 6.
A is lower triangular, so its eigenvalues are its diagonal entries 1, 2 and 4. If \(\lambda\) is an eigenvalue of A, then \(p(\lambda)\) is an eigenvalue of \(p(A)\) for any polynomial p. With \(p(t) = t^{2} - 3t + 2 = (t - 1)(t - 2)\):
\[ \begin{aligned} p(1) &= 0, \quad p(2) = 0 \\ p(4) &= 16 - 12 + 2 = 6 \end{aligned} \]
Option D (0, 3, 12) evaluates something else, and \(-1, 1, 0\) and \(-2, 4, 18\) do not come from substituting 1, 2, 4 into p.
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