# 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.

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

$$T(a\vec i+b\vec j)=-b\vec i+a\vec j\tag6$$.

• Can we visualize the polarisation identity in some way? – Kishalay Sarkar Feb 3 at 13:17