Let $V$ be the vector space of all continuous functions from $\mathbb{R}$ into $\mathbb{R}$ and let $T\colon V \rightarrow V$ be a linear map defined by $T(f)(x)=\int^{x}_{0}f(t)dt$. How can we prove $T$ has no eigenvalue?
Eigen Value of a Linear Map
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$\begingroup$
linear-algebra
eigenvalues-eigenvectors
1 Answers
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If $\lambda$ was an eigenvalue, and $f$ an eigenvector for $\lambda$, then for each $x\in \Bbb R$, we would have $\int_0^xf(t)dt=\lambda f(x).$ If $\lambda$ was equal to $0$, we would have $f\equiv 0$, which is not allowed. So $\lambda\neq 0$, and since $f$ is continuous, $f$ is $C^1$, as a primitive of a continuous function. So we have $f(x)=\lambda f'(x)$ for all $x$ and $f(0)=0$. This gives $\left(\frac 1{\lambda}f(x)-f'(x)\right)e^{-\frac x{\lambda}}=0,$ hence $f(x)=Ce^{\frac x{\lambda}}$. This gives that $f(0)=C=0$, hence $f\equiv 0$.
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0Thank$s$, It is an elegant answer... – 2012-10-01