6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can we visualize the polarisation identity in some way? $\endgroup$ – Kishalay Sarkar Feb 3 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.