3
$\begingroup$

Let $V$ a $\mathbb{C}$-vector space with inner product $\langle \cdot , \cdot \rangle$ and $f:V\to V$. Show that if $\langle f(v),v\rangle\in \mathbb{R}$ for $v \in V$, then $f=f^∗$.

I was thinking of taking the inner product of $\langle f(v),v \rangle$ but I dont really have any clear idea

$\endgroup$
  • 1
    $\begingroup$ You're not very clear on any of the details here. Is $f$ an endomorphism? What exactly is $f^*$ in this notation? Also you should probably include more details on thoughts you had, things you considered trying, etc. $\endgroup$ – Alex Mathers Jun 29 '15 at 21:48
  • 1
    $\begingroup$ Is $f$ a linear map? and $f^*$ the adjoint operator of $f$. Please clarify ! $\endgroup$ – Alonso Delfín Jun 29 '15 at 21:49
5
$\begingroup$

Note that for all $v \in V$ \begin{align} 0 & =\langle fv,v\rangle-\overline{\langle fv,v\rangle} \\ & =\langle fv,v\rangle-\langle v,fv\rangle\\ &=\langle fv,v\rangle -\langle f^{\star}v,v\rangle \\ &=\langle (f-f^{\star})v,v\rangle \end{align}

So $(f-f^{\star})v=\vec{0}$, hence $f=f^{\star}$.

$\endgroup$
  • $\begingroup$ if you use \langle and \rangle instead of <and > you'll get a better formatting. I edited it for you, I hope you don't mind (you can always rollback if you want) $\endgroup$ – Ivo Terek Jun 29 '15 at 22:02
  • 1
    $\begingroup$ No problem! Thanks for the info. $\endgroup$ – mich95 Jun 29 '15 at 22:03
  • 2
    $\begingroup$ This, of course, assumes that $\langle g(x),x \rangle = 0$ for all $x$ implies that $g = 0$. Note that this is true over complex vector spaces, but not over real vector spaces. In the context of real vector spaces, this only guarantees that $g$ is skew-adjoint. $\endgroup$ – Omnomnomnom Jun 29 '15 at 22:05
  • $\begingroup$ Jup, forgot to mention that! $\endgroup$ – mich95 Jun 29 '15 at 22:10
4
$\begingroup$

It suffices to show that $\newcommand{\ip}[1]{\langle #1 \rangle}$ $\ip{f(x),y} = \ip{x,f(y)}$ holds for every $x,y \in V$.

In order to show that this is the case: note that

  • $\ip{f(x),x} = \ip{x,f(x)}$
  • $\ip{f(y),y} = \ip{y,f(y)}$
  • $\ip{f(x+y),x+y} = \ip{x+y,f(x+y)}$
  • $\ip{f(x + iy),x+iy} = \ip{x+iy,f(x+iy)}$

The rest is a matter of applying the sesqui-linearity of the inner product.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.