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?
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 $\Bbb C$".