Part CCSIR NET June 2025normal-means-diagonalisable-so-the-generalised-kernel-is-just-the-kernel

Normal means diagonalisable so the generalised kernel is just the kernel

Let V be a finite dimensional complex inner product space. For a linear map T : V → V, let T* denote its adjoint. Which of the following statements are true?

A.If trace of TT* is zero, then T = 0.✓
B.Let v ∈ V be such that T*T(v) = 0. Then T(v) = 0.✓
C.Suppose T = T* and N > 1 be an integer. Let v ∈ V be such that $T^{2^{N}} (v)$ = 0. Then T(v) = 0.✓
D.Suppose that TT* = T*T and N > 1 be an integer. Let v ∈ V be such that T^N(v) = 0. Then T(v) = 0.✓

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 hypothesis dropped.

See pricing

50 are analysed free — try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

Normal means diagonalisable so the generalised kernel is just the kernel

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