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I am having trouble determining the eigenvalues and eigenvectors of the operator $Kv(x)= \int_0^1((x+t)v(t)dt$, where the kernel is $k=x+t$. I have tried to solve the equation $Kv(x)=\lambda v(x)$, and I know that I should get two eigenvalues, but I can't seem to find them.

Is there a standard method to finding the lambda's other than solving the equation $Kv(x)=v(x)$?

Thanks for the help!

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1 Answer 1

Note that we can write $$Kv(x)=x\int_0^1v(t)dt+\int_0^1tv(t)dt,$$ so if $v$ is an eigenvector for $\lambda$, then we have $Kv(x)=ax+b=\lambda v(x)$. If $\lambda\neq 0$, then we should have $v(x)=\frac a\lambda x+\frac b\lambda $. We have $\int_0^1v(t)dt=\frac a{2\lambda}+\frac b\lambda$. After having computed $\int_0^1tv(t)dt$, we get a linear system. We want a non-trivial solution.

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