# Tough eigenvalue problem, trying to understand it

• jaejoon89
In summary, The eigenvalues of P are 0 and 1. This is shown by the fact that if λ is an eigenvalue of P, then λ=0 or λ=1. Additionally, it is assumed that v is the eigenvector with associated eigenvalue 0 and thus P(v)=0, making v an element of KerP. However, since P(v)=P^2(v), it also follows that v is an element of ImP. This means that v is an element of both KerP and ImP, which is only possible if v=0. Therefore, 0 is not an eigenvalue of P. On the other hand, for P(v)=P^2(v), it is clear
jaejoon89
Suppose P: V->V s.t. P^2 = P and V = kerP + ImP (actually not just + 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?

-----

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.

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.

That 2nd proof certainly makes sense to me. What you have in P is a projection operator, which projects vectors of V to some subspace of V. Now, if the subspace is V itself, then you cannot find the eigenvalue zero for any vector in V. I think that's what happening here but it's certainly not obvious.

Is this part true... I'm not sure it is:

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

The subspace could be zero too -- i.e. P could be the zero operator.

Is this part true... I'm not sure it is:

But P(v) = P^2(v) = P(P(v)) => v in ImP.
Why do you think it's true? If you can't give a precise algebraic calculation that shows it, then you're right to be skeptical about it.

I was thinking that the condition

$$V = \ker(p) \oplus \mathrm{im}(p)$$

might constrain the subspace to be equal to V but it seems that zero works too.

I was thinking that the statement "P(v) = P^2(v) = P(P(v)) => v in ImP" is true because if P(P(v)) then it maps v so obviously v must be in the image. i thought the rest followed from it and I was thinking the same thing as clamtrox, i.e. that the condition constrained it... hm...

jaejoon89 said:
if P(P(v)) then it maps v
I can't parse this.

Are you trying to say that you canceled one P from each side of the equation
Pv = PPv​
? If so, then that's a problem -- you can only cancel invertible matrices. P's action on all of V is not invertible.

## What is an eigenvalue problem?

An eigenvalue problem is a mathematical problem that involves finding the values (eigenvalues) and corresponding vectors (eigenvectors) that satisfy a particular equation. This equation is known as the characteristic equation and is typically written in matrix form.

## Why is the eigenvalue problem considered tough?

The eigenvalue problem is considered tough because it involves solving for the eigenvalues and eigenvectors of a matrix, which can be a complex and time-consuming process. Additionally, there are many different techniques and methods for solving eigenvalue problems, making it challenging to determine the best approach for a particular problem.

## What are some real-world applications of the eigenvalue problem?

The eigenvalue problem has many applications in fields such as physics, engineering, computer science, and economics. Some examples include using eigenvalues to analyze the stability of a system, finding the optimal portfolio in finance, and performing image and signal processing in computer science.

## How can I approach solving a tough eigenvalue problem?

There are several approaches to solving a tough eigenvalue problem, including using numerical methods such as the power method or the QR algorithm, using analytical techniques such as diagonalization or the Cayley-Hamilton theorem, or using software programs specifically designed for solving eigenvalue problems.

## What are some tips for understanding the eigenvalue problem?

One tip for understanding the eigenvalue problem is to have a strong foundation in linear algebra, as this is the branch of mathematics that deals with vectors and matrices. Additionally, practicing with different examples and techniques can help improve understanding and intuition for solving eigenvalue problems.

Replies
5
Views
894
Replies
24
Views
2K
Replies
3
Views
4K
Replies
5
Views
2K
Replies
3
Views
1K
Replies
5
Views
2K
Replies
12
Views
2K
Replies
9
Views
2K
Replies
5
Views
13K
Replies
5
Views
2K