- #1

jaejoon89

- 195

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Which of the following explanations is right? (1 is an eigenvalue, but is 0 also?) Could somebody please explain?

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First answer:

Suppose that λ is an eigenvalue of P, with eigenvector v; then:

Pv = λv = (P^2)v = (λ^2)v

This gives (as v ≠ 0):

λ(1−λ) = 0 ⇔ λ = 0 or 1

In fact, Pv = v iff v ∈ Im(P), because:

v ∈ Im(P) ⇔ v = Pz ⇒ Pv = (P^2)z = Pz = v

On the other hand, if v = n + z and Pv = v, then Pv = P(n + z) = Pz ⇒ v = Pz.

Other answer:

P^2(v) = P(P(v)) = P(pv) = pP(v) = (p^2)v.

But P^2(v) = P(v) = pv.

Thus p^2 = p => p in {0, 1}.

Where v in V is an eigenvector of P:V -> V, with associated eigenvalue p in F, the ground field (V/F).

If P(v) = 0, then v in KerP. (Here we assume that v is the eigenvector with associated eigenvalue 0.)

But P(v) = P^2(v) = P(P(v)) => v in ImP.

Thus v in KerP intersection ImP.

But V = KerP (+) ImP => KerP intersection ImP = {0}.

Thus v = 0 => v is not a valid eigenvector (by convention).

Thus 0 is not an eigenvalue of P.

Thus 1 is the only eigenvalue of P.