Part CCSIR NET December 2025vanishing-at-the-eigenvalues-as-numbers-is-not-the-same-as-being-a-multiple-of-the-minimal-polynomial

Vanishing at the eigenvalues as numbers is not the same as being a multiple of the minimal polynomial

Let A be a non-zero 3×3 matrix with integer entries. Let $\lambda_{i}\in\mathbb{C}, 1\le{}i\le3$ be all the eigenvalues of A (not necessarily distinct). Which of the following statements are necessarily true?

A.There exists a cubic polynomial $f(X)\in\mathbb{Q}[X]$ such that $f(\lambda_{i})=0$ for all 1≤i≤3.✓
B.There exists a quadratic polynomial $f(X)\in\mathbb{Q}[X]$ such that $f(\lambda_{i})=0$ for all 1≤i≤3.
C.If $f(X)\in\mathbb{Q}[X]$ is such that $f(\lambda_{i})=0$ for all 1≤i≤3, then f(A)=0.
D.If $f(X)\in\mathbb{Q}[X]$ is a cubic polynomial such that $f(\lambda_{i})=0$ for all 1≤i≤3, then f(A)=0.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: Same characteristic polynomial ⇒ similar

Vanishing at the eigenvalues as numbers is not the same as being a multiple of the minimal polynomial

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