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

Irreducible quadratics leave no real eigenvector

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

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

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.

See pricing

50 are analysed free — try those first.

Related counterexample: Same characteristic polynomial ⇒ similar

Irreducible quadratics leave no real eigenvector

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