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Part CCSIR NET December 2025non-negative-quietly-admits-zero-which-breaks-the-strict-positivity-the-diagonalizability-argument-actually-needs

Non negative quietly admits zero which breaks the strict positivity the diagonalizability argument actually needs

Let V be a finite-dimensional R−\mathbb{R}-vector space and T:V→V a linear operator such that T2T^{2} is diagonalizable over R\mathbb{R}. Which of the following statements are necessarily true?

  1. A.If T is not diagonalizable over R\mathbb{R}, then T2T^{2} has an eigenvalue ≤0.✓ Correct
  2. B.If T2T^{2} has only negative eigenvalues, then dim V is an even integer.✓ Correct
  3. C.If T2T^{2} has only non-negative eigenvalues, then T is diagonalizable over R\mathbb{R}.
  4. D.For each non-zero v∈V, {v,Tv,T2vT^{2}v} is linearly dependent.

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 boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Related counterexample: Diagonalisable ⇒ invertible

More on this topic

The chapter behind this: Diagonalisability — the criteria card — free to read

From Eigenvalues and Canonical Forms › Diagonalisability criteria

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