# Prove if for a Linear Transformation $T : V \rightarrow V$ over $\mathbb C$ $<v, T(v)> = 0$ for all $v \in V$ then $T(v) = 0$ for all $v \in V$

Prove that if for a Linear Transformation $$T: V \rightarrow V$$ over $$\mathbb C$$ $$\langle v, T(v)\rangle = 0$$ for all $$v \in V$$ then $$T(v) = 0$$ for all $$v \in V$$

Since $$T(v)$$ is also in $$\mathbb C$$ I checked the linear property of two complex numbers to equal zero.

so $$(a + bi)(t + si) = 0$$ would mean that $$asi = -bti$$ and $$at = bs$$ so $$-t^2 = s^2$$ Where t and s are the coefficients of one of the elements in $$T(v)$$ so the only number that equals it's negative is 0.

Hi all. This was a homework assignment and I've come up with an alternative proof (that is probably wrong) to the official (which is simple and I understand, but it's completely different), if it is indeed wrong could someone hint where I'm going in the wrong direction with my thinking?

• It may seem something is missing on your question title. Also it could help if you post the question in the body too Commented Jan 14, 2020 at 8:14
• Did you mean <v,T(v)>=0? Commented Jan 14, 2020 at 8:20
• And what is the hypothesis on that inner product? As it is stated, there is no hypothesis Commented Jan 14, 2020 at 8:23
• You say "<v,T(v)> for all $v\in V$" but you don't specify any property of <v, T(v)> Commented Jan 14, 2020 at 8:33
• @AleTolcachier Oh dear! thank you! Yes, I missed = 0. remidied. So sorry! Commented Jan 14, 2020 at 8:50

From the comments: there has been some confusion over the terminology. $$T$$ is "linear over $$\Bbb C$$" does not mean that the output of $$T$$ is linear. Rather, when we say that $$T:V \to V$$ is a linear transformation "over $$\Bbb C$$", we mean that $$T(\alpha v + \beta w) = \alpha T(v) + \beta T(w)$$ for any complex numbers $$\alpha,\beta$$. So for instance, the conjugation map $$a + bi \mapsto a - bi$$ is a map from $$\Bbb C$$ to $$\Bbb C$$ that is "linear over $$\Bbb R$$" but not "linear over $$\Bbb C$$".