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- Feb 5, 2012

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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.

So we have to find functions such that,Find eigenvectors and eigenvalues of the following linear transformation.

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

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

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

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....

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

\[\Rightarrow A=0\]

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?