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I have some reservations concerning the following proof is it correct ? if not please do point at any mistakes.

NOTE: $\mathcal{L}(\mathbf{C}^\infty)$ denotes the set of all Linear transformations from $\mathbf{C}^\infty$ to $\mathbf{C}^\infty$.

Theorem. The Linear operator $T\in\mathcal{L}(\mathbf{C}^\infty)$ defined by $T(z_1,z_2,...) = (0,z_1,z_2,...)$ has no eigenvalues.

Proof. Assume on the contrary that for some $\lambda\in\mathbf{C}$ there exists a non zero vector $v = (z_1,z_2,...)$ such that $Tv = \lambda v$. Since $v\neq 0$ it follows that the set $\mathcal{N} = \{z_i\neq 0|i\in\mathbf{Z^+}\}$ is non-empty this together with the definition of $T$ implies that $\lambda\neq 0$.

We now show by recourse to Mathematical-induction that $v=0$. Writing out $Tv = \lambda v$ in full we have the following equation. $$T(z_1,z_2,...) = (0,z_1,z_2,...) = \lambda(z_1,z_2,...) = (\lambda z_1,\lambda z_2,...)$$ For $n=1$ it is evident that $z_1 = \frac{0}{\lambda_1} = 0$. Now assume for an arbitrary $k\in\mathbf{Z^+}$ that $z_k = 0$, from the equation above $z_k = \lambda z_{k+1}$ consequently since $\lambda\neq 0$ it follows that $z_{k+1} = \frac{z_k}{\lambda} = \frac{0}{\lambda} = 0$, implying that $v = 0$ contradicting the fact that $v$ is an eigenvector of $T$.

$\blacksquare$

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  • $\begingroup$ Apparently for you $\;\Bbb C^\infty\;$ is the set of all complex sequences. Is this correct? $\endgroup$ – DonAntonio Feb 3 '18 at 13:43
  • $\begingroup$ @DonAntonio Yes Exactly $\endgroup$ – Atif Farooq Feb 3 '18 at 13:44
  • $\begingroup$ @At Thanks. Then I think your proof is correct. $\endgroup$ – DonAntonio Feb 3 '18 at 13:45

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