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Part CCSIR NET December 2024irreducible-quadratics-leave-no-real-eigenvector

Irreducible quadratics leave no real eigenvector

Let T:R4→R4T : \mathbb{R}^{4} \to \mathbb{R}^{4} be an R−\mathbb{R}-linear transformation. Assume that the characteristic polynomial of T has two distinct monic irreducible quadratic factors q1(x)q_{1}(x) and q2(x)q_{2}(x). Which of the following statements are true?

  1. A.There exists a nonzero v∈R4v \in \mathbb{R}^{4} such that q1(T)v=0q_{1}(T)v = 0.✓ Correct
  2. B.There exists a nonzero v∈R4v \in \mathbb{R}^{4} such that q1(T)v=0q_{1}(T)v = 0 and q2(T)v=0q_{2}(T)v = 0.
  3. C.For all nonzero v∈R4,vv \in \mathbb{R}^{4}, v and Tv are linearly independent.✓ Correct
  4. D.If q1(T)v=0q_{1}(T)v = 0 for some nonzero v∈R4v \in \mathbb{R}^{4}, then {v, Tv, T2v,T3vT^{2}v, T^{3}v} is a basis of R4\mathbb{R}^{4}.

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 base field or ring.

See pricing

50 are analysed free — try those first.

The trap it tests

Base field or ring

The answer changes with the field or ring you are working over.

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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