Correct answer: (A) Eigenvalue 3 has multiplicity 2, with only one independent eigenvector.
Explanation
The characteristic polynomial is $(5-\lambda)(1-\lambda)+4=\lambda^2-6\lambda+9=(\lambda-3)^2$. At $\lambda=3$, $A-3I$ has rows $(2,-1)$ and $(4,-2)$, so its rank is 1 and its null space has dimension 1. Thus the repeated eigenvalue has only one independent eigenvector, for example $(1,2)^T$. Select A.
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