# Show $f=f^*$ including inner product

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

• 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. – Alex Mathers Jun 29 '15 at 21:48
• Is $f$ a linear map? and $f^*$ the adjoint operator of $f$. Please clarify ! – Alonso Delfín Jun 29 '15 at 21:49

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 $f=f^{\star}$.

• 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) – Ivo Terek Jun 29 '15 at 22:02
• No problem! Thanks for the info. – mich95 Jun 29 '15 at 22:03
• 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. – Omnomnomnom Jun 29 '15 at 22:05
• Jup, forgot to mention that! – mich95 Jun 29 '15 at 22:10

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.