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Part CCSIR NET June 2024minimal-polynomial-gives-the-eigenvalues

Minimal polynomial gives the eigenvalues

Let T:R4→R4T : \mathbb{R}^{4} \to \mathbb{R}^{4} be a linear map with four distinct eigenvalues and satisfying T4−15T2+10T+24I=0T^{4} - 15T^{2} + 10T + 24I = 0. Which of the following statements are necessarily true?

  1. A.There exists a non-zero vector v1∈R4v_{1} \in \mathbb{R}^{4} such that Tv1=2v1_{1} = 2v_{1}✓ Correct
  2. B.There exists a non-zero vector v2∈R4v_{2} \in \mathbb{R}^{4} such that Tv2=v2_{2} = v_{2}
  3. C.For every non-zero vector v∈R4v \in \mathbb{R}^{4}, the set {2v, 3Tv} is linearly independent
  4. D.T is a one-one function✓ Correct

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 standard counterexample.

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50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

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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