Find the eigenvalues of $\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$.
A2 and 3
B1 and 0
C-2 and -3
D5 and 6
Answer & Solution
Correct answer: A. 2 and 3
1. For a TRIANGULAR (upper or lower) matrix, the eigenvalues are exactly the DIAGONAL entries.
2. This matrix is upper triangular with diagonal entries 2 and 3.
3. So eigenvalues are 2 and 3.
4. Verify: $\det(A - \lambda I) = (2-\lambda)(3-\lambda) = 0 \Rightarrow \lambda = 2, 3$. ✓
5. Sanity: trace = 5 = 2+3 ✓; det = 6 = 2·3 ✓.
6. Other options misidentify the eigenvalues.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.5 (Triangular matrices — eigenvalues are diagonal)._
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