Suppose P: V->V s.t. P^2 = P and V = kerP + ImP (actually notjust + but a direct

Question

Suppose P: V->V s.t. P^2 = P and V = kerP + ImP (actually notjust + but a direct sum). Find all eigenvalues of P.

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

-----
First answer:

Suppose that is an eigenvalue of P, with eigenvector v;then:
Pv = v = (P^2)v = (^2)v
This gives (as v 0):
(1) = 0 = 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 associatedeigenvalue p in F, the ground field (V/F).
If P(v) = 0, then v in KerP. (Here we assume that v is theeigenvector 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.

Explanation / Answer

Your response that the eigenvalues are both 1 and 0 iscorrect. The value of zero for the eigenvalue is still important andworth mentioning. Your response that the eigenvalues are both 1 and 0 iscorrect. The value of zero for the eigenvalue is still important andworth mentioning.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.