Linear Transformation with No Eigenvector

[Sudharaka]

Hi everyone, :)

This is one of those questions I encountered when trying to do a problem. I know that a eigenvector of a linear transformation should be non-zero by definition. So does that mean every linear transformation has eigenvectors? What if there's some linear transformation where no eigenvectors exist. Here's a question which I came across which motivated me to think about this.

Find eigenvectors and eigenvalues of the following linear transformation.

\[T:\,f(x)\rightarrow\int_{0}^{x}f(t)\,dt\] in the linear span over \(\Re\).

So we have to find functions such that,

\[T(f(x))=\lambda\,f(x)\]

\[\Rightarrow \int_{0}^{x}f(t)\,dt=\lambda \,f(x)\]

Now differentiating both sides we get,

\[f(x)=\lambda\,f'(x)\]

Solving for \(f\) we finally obtain,

\[f(x)=Ae^{x/\lambda}\]

where \(A\) is a constant. To find the value of \(A\) we plug this to our original equation. And then something out of the ordinary happens... :p

\[\int_{0}^{x}Ae^{t/\lambda}\,dx=\lambda\,Ae^{x/\lambda}\]

\[\Rightarrow A=0\]

So we get, \(f(x)=0\) which is not a eigenvector. And hence I came to the conclusion that this linear transformation doesn't have any eigenvectors. Am I correct? :)

 

[Ackbach]

\[\int_{0}^{x}Ae^{t/\lambda}\,dx=\lambda\,Ae^{x/\lambda}\]

\[\Rightarrow A=0\]

Why does $\lambda \, A e^{x/ \lambda}=\lambda \, A e^{x/ \lambda}$ require $A=0$?

 

[Sudharaka]

Why does $\lambda \, A e^{x/ \lambda}=\lambda \, A e^{x/ \lambda}$ require $A=0$?

Thanks for the reply. Well if I am not wrong here, I think you have calculated the integration incorrectly. :)

\[\int_{0}^{x}Ae^{t/\lambda}\,dx=\lambda\,Ae^{x/\lambda}\]

\[\Rightarrow \lambda \, A e^{x/ \lambda}-\lambda\,A=\lambda\,Ae^{x/\lambda}\]

\[\therefore A=0\]

 

[Ackbach]

Thanks for the reply. Well if I am not wrong here, I think you have calculated the integration incorrectly. :)

\[\int_{0}^{x}Ae^{t/\lambda}\,dx=\lambda\,Ae^{x/\lambda}\]

\[\Rightarrow \lambda \, A e^{x/ \lambda}-\lambda\,A=\lambda\,Ae^{x/\lambda}\]

\[\therefore A=0\]

Oh, you're right. I think I misinterpreted your equation as saying that's what the integral was; I didn't bother to "check" it.

Yeah, if the exponential function doesn't work, nothing will. Thinking of it in terms of series expansions, the integration operator will always raise the power by one. So it seems unlikely that the resulting series would be a multiple of the original.

 

[Opalg]

I came to the conclusion that this linear transformation doesn't have any eigenvectors. Am I correct? :)

Yes, you are correct. See here.

 

[Sudharaka]

Yes, you are correct. See here.

Thanks so much. :) I didn't know that the linear transformation is called the Volterra operator and was confused by the fact that it didn't have any eigenvalues. I have never encountered such a thing before. Thanks again for your valuable input. :)

 

[HallsofIvy]

If V is a vector space over the complex numbers then any linear transformation has eigenvalues and so non-zero eigenvectors. If V is a vector space of the real numbers it may not have eigenvalues (solutions to the characteristic equation are all complex) and so no eigenvectors.

Yes, the standard definition of "eigenvector" in most texts require that it be non-zero. I personally don't like that and prefer, as you will see in some texts:
"\lambda is an eigenvalue for linear transformation A if and only if there exist a non-zero vector, v, such that Av= \lambda v."

"An eigenvector, corresponding to eigenvalue \lambda is any vector, v, such that Av= \lambda v"

That allows the 0 vector as an eigenvector (for any eigenvalue) so that we can say "The set of all eigenvectors corresponding to eigenvalue \lambda form a vector space" rather than having to say "The set of all eigenvectors corresponding to eigenvalue \lambda, together with the 0 vector, form a vector space"

A very minor detail either way!

 

[I like Serena]

If V is a vector space over the complex numbers then any linear transformation has eigenvalues and so non-zero eigenvectors. If V is a vector space of the real numbers it may not have eigenvalues (solutions to the characteristic equation are all complex) and so no eigenvectors.

I can't find any complex eigenvalues for this particular problem...

 

[Opalg]

If V is a vector space over the complex numbers then any linear transformation has eigenvalues and so non-zero eigenvectors. If V is a vector space of the real numbers it may not have eigenvalues (solutions to the characteristic equation are all complex) and so no eigenvectors.

I can't find any complex eigenvalues for this particular problem...

HallsofIvy's comments refer to linear transformations on finite-dimensional vector spaces. The integral operator in this thread acts on a space of functions that is infinite-dimensional, and the theory is quite different in that situation. A continuous linear operator on a normed vector space has associated with it a set of complex numbers called its spectrum. If the space is finite-dimensional then the spectrum consists exactly of the eigenvalues. In the infinite-dimensional case, the spectrum is always a nonempty set, but it does not necessarily consist of eigenvalues. For the Volterra integral operator, the spectrum consists of the single point $\{0\}$ (which is not an eigenvalue).

 

[HallsofIvy]

Yes, thank you. My mind tends to go foggy when I think about infinite dimensional vector spaces so I just ignore them!