Proving that a transformation is Unitary

Question

Let $T:V\rightarrow V$ be a linear, normal transformation (meaning $TT^*=T^*T$) where it is known that $T^{-1} = -T$. Can it be proved that $T$ is unitary (e.g. $TT^*=T^*T=I$)?

Answer 1

If $V$ is finite-dimensional, we have that there is an orthonormal basis $(e_i)$ of $V$ consisting of eigenvectors of $T$, say $Te_i = \lambda_i e_i$. We have $T^*e_i = \bar\lambda_i e_i$ and $T^{-1}e_i = \lambda_i^{-1}e_i$. As $T^{-1} = -T$, we have $\lambda_i^{-1} = -\lambda_i$ or $\lambda_i^2 = -1$ for each $i$. Hence $\lambda_i = \pm \imath$ for every $i$. Now, for each $i$, we have $$ TT^*e_i = \lambda_i\bar\lambda_i e_i = (\pm\imath)(\mp \imath)e_i = e_i $$ Hence, $TT^* = \mathrm{Id}$, as both operators agree on a basis.

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For infinite dimensional, but complete $V$, we can argue along the same line: As $T^{-1} = -T$, or $T^2 + \def\Id{\mathrm{Id}} = 0$, we have $\sigma(T) \subseteq \{\pm \imath\}$, so there is a spectral measure supported in $\{\pm \imath\}$, such that $$T = \int_{\pm \imath}\lambda\, dE(\lambda) = \imath E(\imath) -\imath E(-\imath)$$ Now $$ T^* = -\imath E(\imath) + \imath E(-\imath) $$ Hence, \begin{align*} TT^* &= (-\imath)^2 E(\imath) + \imath^2 E(-\imath)\\ &= E(\imath) + E(-\imath)\\ &= E\bigl(\sigma(T)\bigr)\\ &= {\mathrm Id} \end{align*}