Prove that there is no non-zero linear operator on $\mathbb C^{2}$ such that $(\alpha | T\alpha) = 0$ I'm stuck on this problem for a long time. I can't find a proper solution for this without using complex calculations. I hope some one can help me with this problem:


Let $ (\;|\;)$ be the standard inner product on $\mathbb C^{2}$. Prove that there is no non-zero linear operator $T$ on $\mathbb C^{2}$ such that $(\alpha|T\alpha) = 0\;$ for every $\alpha$ in $\mathbb C^{2}$. Generalize.


 A: This is not possible in $\mathbb{C}^2$ and this is slightly tricky.
Note that the links provided by Jonas Meyer show you the most straighforward route: if $(Tx,x)=0$ for all $x$, then $(Tx,y)=0$ for al $x,y$ so $T=0$.
Now here is another approach which is an opportunity to see two important facts about bounded linear operator on a complex Hilbert space.
Claim 1: a bounded operator on a complex Hilbert space is self-adjoint ($T^*=T$) if and only if $(Tx,x)\in\mathbb{R}$ for all $x$. 
Remark: this fails in the real case as shows the example down below. Note the use of $iy$ and semilinearity of the inner product in the last step of the proof.
Proof: If $T^*=T$ first then
$$
(Tx,x)=(x,T^*x)=(x,Tx)=\overline{(Tx,x)}
$$
is real.
Conversely, we will prove that $(x,Ty)=(Tx,y)$ for all $x,y$, which is clearly equivalent to $T$ being self-adjoint since $(Tx,y)=(x,T^*y)$.
Since $(Tx+Ty,x+y)$ is real, we have
$$
(Tx+Ty,x+y)=\overline{(Tx+Ty,x+y)}=(x+y,Tx+Ty).
$$
Expanding the lhs and the rhs yields
$$
(Tx,y)+(Ty,x)=(x,Ty)+(y,Tx).
$$
Applying this to $x$ and $iy$ yields
$$
(Tx,y)-(Ty,x)=(x,Ty)-(y,Tx).
$$
Summing these two equalities, we get $(Tx,y)=(x,Ty)$ as desired. 
End of the proof.
Claim 2: If $T$ is a self-adjoint bounded operator on a Hilbert space, then
$$
\|T\|=\sup_{\|x\|=1}|(Tx,x)|.
$$
Proof: the inequality $\geq$ is trivial. The other direction use the fact that 
$$
\|T\|=\sup_{\|x\|=1}\|Tx\|=\sup_{\|x\|=\|y\|=1}|(Tx,y)|
$$
which follows directly from the fact that $\|z\|=\sup_{\|y\|=1}|(x,y)|$, a simple consequence of Cauchy-Schwarz.
From here, you can use the polarization identity and the parallelogram law to conclude. But I will only give details if you ask since this is already pretty long.
End of proof.
As an application of the above claims, if $(Tx,x)=0$ for all $x$, then $T$ is self-adjoint so
$$
\|T\|=\sup_{\|x\|=1}|(Tx,x)|=0,
$$
hence $T=0$.
So such a nonzero $T$ does not exist.
But it is possible in $\mathbb{R}^2$: $$\left(\matrix{0& 1\\-1&0} \right).$$
A: Claim 1:  If for two complex linear operators $T,S$ we have $\langle{Tx,x \rangle}=\langle{Sx,x \rangle}\forall x \in \mathbb C^{2}$ then $\langle{Tx,y \rangle}=\langle{Sx,y \rangle} \forall x,y \in \mathbb C^{2}$ (analogous result holds with result to linear operator being in the second argument also)
Proof: This has been answered here
Claim 2: If for a complex linear operator $T$ we have $\langle{x,Ty \rangle}=0 \forall x,y \in \mathbb C^{2}$ then $T \equiv O$ (zero operator).
Proof: Put $x=Ty$, then $\langle{x,Ty \rangle}=\langle{Ty,Ty \rangle}=0$
$\implies Ty=0 \in \mathbb C^{2}$
$\implies T \equiv O$
Main Proof: $\langle{x,Tx \rangle}=0=\langle{x,Ox \rangle},\forall x \in \mathbb C^{2}$ ($O$ is the zero operator)
$\implies \langle{x,Ty \rangle}=\langle{x,Oy \rangle}=\langle{x,O \rangle}=0 \forall x,y \in \mathbb C^{2}$(from claim 1)
$\implies T \equiv O$ (from claim 2)
So no non-zero complex linear operator satisfies the given condition.
