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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 λi∈C,1≤i≤3\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?

  1. A.There exists a cubic polynomial f(X)∈Q[X]f(X)\in\mathbb{Q}[X] such that f(λi)=0f(\lambda_{i})=0 for all 1≤i≤3.✓ Correct
  2. B.There exists a quadratic polynomial f(X)∈Q[X]f(X)\in\mathbb{Q}[X] such that f(λi)=0f(\lambda_{i})=0 for all 1≤i≤3.
  3. C.If f(X)∈Q[X]f(X)\in\mathbb{Q}[X] is such that f(λi)=0f(\lambda_{i})=0 for all 1≤i≤3, then f(A)=0.
  4. D.If f(X)∈Q[X]f(X)\in\mathbb{Q}[X] is a cubic polynomial such that f(λi)=0f(\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, and why this one tests invariants don't determine.

See pricing

50 are analysed free — try those first.

The trap it tests

Invariants don't determine

Two objects sharing an invariant were treated as the same object.

Drill statements like this

Related counterexample: Same characteristic polynomial ⇒ similar

More on this topic

The chapter behind this: Characteristic vs minimal polynomial — what each tells you — free to read

From Eigenvalues and Canonical Forms › Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

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