Prove that if $\langle T(v),v\rangle=0$ for every $v\in V$ then $T=0$. Show that the same result is not true over $\mathbb{R}$

Question

Let $V$ be a vector space over $\mathbb{C}$ and $T\in L(V)$. Prove that if $\langle T(v),v\rangle=0$ for every $v\in V$ then $T=0$. Give an example to show that the same result is not true if $V$ is a vector space over $\mathbb{R}$.

My attempt:

I first thought doing the following:

If $\langle T(v),v\rangle=0$ for every $v\in V$ that means $T(v)\in V^{\perp}=\{0\} $ , then $T(v)=0$ for every $v\in V$ and then $T=0$

Second, I tried another proof:

If $\langle T(v),v\rangle=0$ for every $v\in V$, then $\langle T(v),v\rangle=\langle v,T(v)\rangle$ what means $T$ is self adjoint $(T=T^*)$ and positive, then, if the dimension of the space is finite, there exist $F\in L(V)$ invertible such that $T=F^*F$ but in this case:

$\langle T(v),v\rangle=\langle F^*F(v),v\rangle=\langle F(v),F(v)\rangle=\lVert F(v)\rVert^2=0$ then $F(v)=0$ and $F=0$ what means $T=0$

How can I do a general proof that doesn't require finite dimensional vector space?

If $V$ is a vector space over $\mathbb{R}$, we can consider

$[T]_{can}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$ and in this case, $\langle T(e_1),e_1\rangle=0$ and $\langle T(e_2),e_\rangle=0$ but $T\neq 0$

What am I doing wrong?

Edit:

As everything I did before doesn't work, I tried to think that if V is a vector space over $\mathbb{C}$ then we can find a eigenvalue $\lambda$ with eigenvector $v$ so, we have $0=\langle T(v),v\rangle=\langle\lambda v,v\rangle=\lambda\langle v,v\rangle=\lambda ||v||$ and $\lambda=0$ because $v\neq 0$, we can conlude then that every eigenvalue has to be zero but I don't know how to continue here. What happens when a operador has only zero eigenvalues?

Answer 1

You are doing several things wrong, unfortunately.

1. The condition $\langle T(v),v\rangle=0$ does not mean that $T(v)\in V^{\perp}$ for all $v$: you are not saying that $\langle T(v),x\rangle=0$ for all $x$, you are only saying that $T(v)$ is perpendicular to $v$. So your first thought is incorrect. You cannot conclude that $T(v)\in V^{\perp}$.

2. Same problem with your second attempt. For $T$ to be self-adjoint you need $\langle T(x),y\rangle = \langle x,T(y)\rangle$ for all $x$ and $y$, not just when $x=y$. So you cannot conclude from $\langle T(v),v\rangle = \langle v,T(v)\rangle$ for all $v$ that $T$ is self-adjoint.

3. Your example doesn't work: note that $T(1,1) = (1,1)$, so $\langle T(1,1),(1,1)\rangle = 2\neq 0$. Just because each basis vector maps to a vector that is orthogonal to it does not mean that every vector is mapped to a vector that is orthogonal to it.
   We can check a property just on a basis when we have some kind of linearity. But here what we have is an assignment that sends the vector $v$ to the scalar $\langle T(v),v\rangle$. This is not linear: because $v+w$ is mapped to $$\langle T(v+w),v+w\rangle = \langle T(v),v\rangle + \langle T(v),w\rangle + \langle T(w),v\rangle + \langle T(w),w\rangle,$$ and not to $\langle T(v),v\rangle+\langle T(w),w\rangle$. So checking the condition on a basis does not allow you to conclude it holds at every vector.
   But maybe if instead of a reflection (your example), you try a rotation?

Note that neither of your attempted proofs ever uses the fact that you are working over $\mathbb{C}$, so there must surely be something wrong in them.

Answer 2

Consider for example for $T$ in the plane a rotation of $\pi/2$ then $\langle T(v),v\rangle=0$ for all $v$. $T$ is not $0$.

Answer 3

we have $$\langle Tx,y\rangle =\frac{1}{4} (\langle T(x+y),x+y\rangle-\langle T(x-y),x-y\rangle+\langle T(x+iy),x+iy\rangle-\langle T(x-iy),x-iy\rangle)$$,So if $\forall v\in V,\langle T(v),v\rangle=0$, then we conclude that $\forall x,y \in V;\langle Tx,y\rangle =0$, in particular for $y=Tx$, so $T=0$.