Discussion
Let \(A\) be a \(3 \times 3\) real matrix having eigenvalues 1, 2, and 3. If \(B = A^{2} + 2A + I\), where \(I\) is the \(3 \times 3\) identity matrix, then the eigenvalues of \(B\) are
Answer: A
The eigenvalues of \(B\) are 4, 9, 16.
If \(Av = \lambda v\), then \(A^{2}v = \lambda^{2}v\), so any polynomial in \(A\) has the same eigenvector with the polynomial applied to the eigenvalue:
\[ Bv = (\lambda^{2} + 2\lambda + 1)v = (\lambda + 1)^{2} v \]
For \(\lambda = 1, 2, 3\) this gives \(2^{2}, 3^{2}, 4^{2} = 4, 9, 16\).
1, 2, 3 are the eigenvalues of \(A\) itself, 1, 4, 9 are those of \(A^{2}\) alone, and 4, 16, 25 squares \(\lambda + 1\) wrongly shifted for the last two.
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