Can someone check if the following argument is valid? Show that if $T$ is an operator and operator $T \circ T$ has eigenvalue $\lambda^{2}$, than $\lambda$ or $-\lambda$ is an eigenvalue of $T$. $\textbf{Proof.}$ Let $\overline{x}$ be eigenvector corresponding to eigenvalue $\lambda$. From the assumtion we have that $T^{2}\overline{x}=\lambda^{2}\overline{x},$ so $T^{2}\overline{x}-\lambda^{2}\overline{x}=0,$ then $0=(T^{2}-\lambda^{2})\overline{x}.$ From the assumption that $\overline{x}$ is an eigenvector we have $\overline{x}\neq \overline{0}$. This implies that $(T^{2}-\lambda^{2})=0$, but $0=(T^{2}-\lambda^{2})=(T-\lambda)(T+\lambda)=(T-\lambda)(T-(-\lambda)).$ Therefore $ (T-\lambda)=0\ or\ (T-(-\lambda))=0,$ so $\lambda$ or $-\lambda$ is an eigenvalue of the operator T. $\blacksquare$
Is this argument valid?
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operator-theory
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0@RobertIsrael Thanks! – 2012-11-13
1 Answers
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Not quite. By writing $(T^2-\lambda^2)=0$, you are making the claim that it is the $0$ operator. This is not true. Your factorization does work since $\lambda T(\bar{x})=T(\lambda\bar{x})$. So, what you know is that $ (T^2-\lambda^2)(\bar{x})=(T-\lambda)(T+\lambda)(\bar{x}). $ You can still argue from there.
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0Thanks, that solves my problem! – 2012-11-13