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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. $\blacksquare$

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    It's not true that $\bar{x} \ne 0$ implies $T^2 - \lambda^2 = 0$. There are some nonzero vectors in the kernel of that operator, but not all are.2012-11-13
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    Fortunately, you don't need that. $(T^2 - \lambda^2) \overline{x} = (T - \lambda) (T + \lambda) \overline{x} = 0$ implies that either $(T+\lambda) \overline{x} = 0$ or $y = (T+\lambda)\overline{x}$ is a nonzero vector with $(T-\lambda) y = 0$.2012-11-13
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    @RobertIsrael Thanks!2012-11-13

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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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    Thanks, that solves my problem!2012-11-13