Visualization of a "Not so intuitive" problem of linear algebra. I recently encountered a problem in Hoffmann-Kunze linear algebra:
If $(.,.)$ is the standard inner product on $\mathbb C^2$ then show that $(Tv,v)=0 \forall v\in \mathbb C^2 \implies T=0$, I think the proof is quite tricky and I saw its solution at last after trying myself through a long time. The solution is like this, we put $x+y$ and $x+iy$ for $v$ and prove that $(Tx,y)=0 \forall x,y\in \mathbb C^2$, thus it follows that $Tx=0 \forall x\in \mathbb C^2$, but I think it cannot be just a trick, there might be a deeper understanding within this simple 'just a trick' looking proof. I want to visualize why this fact is not true for $\mathbb R^2$ space but true when $\mathbb C$ replaces $\mathbb R$. 
I am looking for an intuitive solution of the same problem or a suitable visualization that would help me in better understanding of this problem.
 A: The polarization identity allows us to look at this in terms of distances. We have 
$\langle Tv,v\rangle=\frac{1}{4}(\|Tv+v\|^2-\|Tv-v\|^2+i\|Tv+iv\|^2-i\|Tv-iv\|^2)\tag1$ 
and if this is to be equal to zero, then 
$\|Tv+v\|^2=\|Tv-v\|^2 \tag2$
$$\text{and}$$ 
$\ \|Tv+iv\|^2=\|Tv-iv\|^2\tag3$.
Suppose $v=(1,0)$. Then, from $(2),$ we see that $Tv$ moves $v$ to a point equidistant from $(1,0)$ and $(-1,0)$, so $Tv$ must be pure imaginary. On the other hand, from $(3),$ we have that $T$ moves $v$ to a point equidistant from  $(0,1)$ and $(0,-1)$ so $T$ must be pure real. It follows that $Tv=0$ and hence by linearity, that $T$ maps the $x$-axis to zero. Similarly, with $v=(0,1)$, the same reasoning shows that $T$ maps the $y$-axis to zero. It follows that $T=0.$
In $\mathbb R^2,$ we have 
$\langle Tv,v\rangle=\frac{1}{4}(\|Tv+v\|^2-\|Tv-v\|^2)\tag4$ 
so if this is zero, then it's only necessary that 
$\|Tv+v\|^2=\|Tv-v\|^2\tag5$
and it's easy to cook up non-zero transformations that satisfy this. For example, a $90$-degree rotation: for $v=a\vec i+b\vec j;\ a,b\in \mathbb R$, set
$T(a\vec i+b\vec j)=-b\vec i+a\vec j\tag6$.
