Let $V$ be an inner product space and $ \beta, \gamma$ fixed vectors in $V$. Show that $T \alpha = (\alpha\mid\beta) \gamma$ defines a linear operator on $V$. Show that $T$ has an adjoint, and describe $T^*$ explicitly.


The Linearity is easy. The problem is to find $T^*$.

$$ (Tv\mid u)=((v\mid \beta) \gamma\mid u)= (v\mid\beta)(\gamma\mid u)=(v\mid\overline{(\gamma\mid u)}\beta)$$.

Then $T ^ * u = \overline{( \gamma\mid u)} \beta = {(u\mid\gamma)}\beta$.

Is correct? I'm not sure you can do this $((v\mid\beta)\gamma\mid u)= (v\mid\beta)(\gamma\mid u)$. Any suggestion?

  • 1
    $\begingroup$ That's perfectly fine, as $(v \mid \beta)$ is a scalar like any other. The answer is correct. $\endgroup$ – Theo Bendit Feb 17 at 13:25
  • $\begingroup$ @TheoBendit Thanks for the comment! $\endgroup$ – Ricardo Freire Feb 17 at 13:27

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