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Give an example of a linear operator T from an infinite dimensional complex vector space X to X such that it has no any eigenvalues.

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The space $X$ must be infinite dimensional. Why?

Take the space $X$ having a countable basis $\{v_n:n\in\mathbb{N}\}$ and the operator $T\colon X\to X$ defined by $$ T(v_{n})=v_{n+1} $$ A nonzero vector in $X$ is of the form $$ v=\alpha_1v_1+\dots+\alpha_nv_n $$ with $\alpha_n\ne0$; then $$ T(v)=T(\alpha_1v_1+\dots+\alpha_nv_n)= \alpha_1v_2+\dots+\alpha_nv_{n+1} $$ which cannot be a scalar multiple of $v$.

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  • $\begingroup$ kindly give some hint to prove that it has no eigen values. $\endgroup$ – Ajit Apr 29 '17 at 16:20
  • $\begingroup$ @Infinity Added some more hints $\endgroup$ – egreg Apr 29 '17 at 16:29
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Integration operator from $C[0,1]$ to $C[0,1]$ has no eigen values.

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