Proof that $\langle x , T(x) \rangle = 0$ implies $T(x) = 0$ for a vector space over a complex field.

Question

I know that this has a proof at Relation between $T=0$ and $(Tx,x)=0$ (and some other questions) but my question is specifically about how to prove this by contradiction rather than directly. (I thought it would be easier this way, but I'm really struggling to come up with a proof).

The exact statement I'm trying to prove is this: let $V$ be a vector space, and $T : V \to V$ a linear transformation. Assume that thist vector space is over the field $\mathbb{C}$, then prove that if $\forall x\Big(\langle x, T(x)\rangle = 0 \Big)$, then $T = 0$.

I started like this

• Assume that the statement is false, then it must be that $\langle x, T(x) \rangle = 0$ and $T \ne T_0$.
• Therefore, there exists $a, b \in V$ such that $b = T(a)$ and $a \ne 0$ and $b \ne 0$. By assumption we have that $\langle a, b \rangle = 0$. From here I think I should try to show that $b=0$, and the fact that this is over a complex field must be important (as the statement is false if we are working over a real field). Unfortunately I got really stuck here. I thought about trying to express $a$ in terms of something like $x + iy$ (for some vectors $x, y$ both with real coordinates) and then to write $b = T(x + iy)$, so that we have

$$ \begin{align} \langle x + iy, T(x + iy) \rangle &= \langle x, T(x) \rangle + \overline{i} \langle x, T(y) \rangle + i \langle y, T(x) \rangle + i \overline{i} \langle y, T(y) \rangle \\ &= 0 + \overline{i} \langle x, T(y) \rangle + i \langle y, T(x) \rangle + 0 \\ &= i\left(- \langle x, T(y) \rangle + \langle y, T(x) \rangle \right) \end{align} $$

• And this must be equal to zero, therefore $- \langle x, T(y) \rangle + \langle y, T(x) \rangle = 0$, that is $\langle x, T(y) \rangle = \langle y, T(x) \rangle$ but I have no idea what to do with this.

Any help would be much appreciated (sorry for the mess).

Answer 1

Your idea is good, but contradiction is not necessary.

I usually have inner products semilinear in the first variable and linear in the second one, fix the argument for your convention, if needed.

First we “polarize”: since $\langle x+y,T(x+y)\rangle=0$ for every $x,y\in V$, we can expand and obtain $$ \langle x,T(y)\rangle+\langle y,T(x)\rangle=0 $$ (write down the steps). This holds for every $x$ and every $y$, so it remains valid when we replace $x$ by $ix$ and we obtain $$ -i\langle x,T(y)\rangle+i\langle y,T(x)\rangle=0 $$ Removing the factor $i$ and summing the two relations, we obtain $$ 2\langle y,T(x)\rangle=0 $$ for every $x,y\in V$. Now you should be able to finish.