Let $f$ be an endomorphism of the (finite-dimensional) unitary space $(V,\langle.,\rangle)$. Show that if $\langle f(x), x\rangle=0$ for all $x\in V$, then $f=0$.
I know this has been posted, however, feel free to help me out with my attempts to prove it.
- Proof
Let $x \in V$, then $f(x) = x$ because it holds for all $x \in V$. We get $\langle f(x), f(x) \rangle=0$ and from there $||f(x)||^2$ = 0 $\rightarrow$ $f(x) = 0$
What am I missing here?
- Proof
Let $x=a+bi$, $a, b \in V$. \begin{align*} 0=\langle f(a+b i), a+b i\rangle & =\langle f(a), a\rangle-i\langle f(a), b\rangle+\langle i f(b), a)+\langle i f(b), b i\rangle \\ & =\langle f(a), a\rangle+\langle f(b), b\rangle+i(-\langle f(b), a\rangle+\langle f(a), b\rangle) \\ & =i(-\langle f(b), a\rangle+\langle f(a), b\rangle) \end{align*} \begin{align*} \Leftrightarrow 0=i(-\langle f(b), a\rangle+\langle f(a), b\rangle) \\ \Leftrightarrow 0=\langle f(a), b\rangle-\langle f(b), a\rangle \\ \Leftrightarrow\langle f(b), a)=\langle f(a), b\rangle \end{align*}
How can I get the job done here? Maybe with defining some $a,b$?
Thanks for your help!
