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Let $V$ be an inner product space and $T$ a linear operator with $T(\alpha) = (\alpha,\beta)\gamma$ for fixed elements $\beta,\gamma \in V$.

I now that $T$ is linear operator. How we can show that adjoint of $T$ ($T^*$) exist and what is it?

SKMohammadi
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1 Answers1

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By the very definition of $T^*$ we must have $$ (T\alpha, \delta) = (\alpha, T^*\delta), \qquad \alpha,\delta \in V $$ So, in our case \begin{align*} (\alpha, T^*\delta) &= (T\alpha, \delta)\\ &= \bigl((\alpha,\beta)\gamma, \delta\bigr)\\ &= (\alpha, \beta)(\gamma,\delta)\\ &= \bigl(\alpha, \overline{(\gamma,\delta)}\beta\bigr)\\ &= \bigl(\alpha, (\delta, \gamma)\beta\bigr) \end{align*} so $T^*\delta = (\delta, \gamma)\beta$, all $\delta \in V$.

martini
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  • thank you @martini, but why $((\alpha,\beta)\gamma,\delta)=(\alpha,\beta)(\gamma,\delta)$? (,) denote inner product!?! :) – SKMohammadi Aug 07 '14 at 08:44
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    The inner product is linear, $(\alpha,\beta)$ is just a scalar, let for a momoent $a := (\alpha, \beta) \in \mathbb C$. Then $$\bigl((\alpha, \beta)\gamma, \delta\bigr) = (a\gamma, \delta) = a(\gamma,\delta) = (\alpha,\beta)(\gamma, \delta) $$ – martini Aug 07 '14 at 08:55