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Part BCSIR NET December 2024the-minimal-polynomial-divides-the-gcd

The minimal polynomial divides the gcd

Let T:R4→R4T : \mathbb{R}^{4} \to \mathbb{R}^{4} be an R−\mathbb{R}-linear transformation such that (T2+T+I)(T−2I)=0(T^{2} + T + I)(T - 2I) = 0 and (T2+T+2I)(T2−4I)=0(T^{2} + T + 2I)(T^{2} - 4I) = 0. Which of the following statements is FALSE?

  1. A.T is diagonalizable over R\mathbb{R}.
  2. B.The characteristic polynomial of T is (x−2)4(x - 2)^{4}.
  3. C.The characteristic polynomial of T is (x2+x+2)(x2−4)(x^{2} + x + 2)(x^{2} - 4).✓ Correct
  4. D.For every R−\mathbb{R}-linear transformation S:R4→R4S : \mathbb{R}^{4} \to \mathbb{R}^{4}, we have ST = TS.

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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