Suppose $T$ is an operator on a complex inner product space $V$ such that $\langle Tv, v \rangle = 0$ for all $v \in V$. Then $T = 0$.
The result above is false for real inner product spaces - a counterexample being the rotation operator.
Are real inner product spaces not a subset of complex inner product spaces?
It seems contradictory to me that a result can only be true for a more general case.
Actually, it's essentially the other way around: intuitively, being a complex inner product space is a more restrictive condition than being a real inner product space.
Let's start by ignoring the inner product structure, since that already reveals the key point:
If I have a vector space $V$ over $\mathbb{C}$, then $V$ can also be viewed as a vector space over $\mathbb{R}$ - since $\mathbb{R}\subseteq\mathbb{C}$, I can just restrict the scalar multiplication in the $\mathbb{C}$-sense to real scalars. The converse, however, is false: there's no obvious way to take a vector space $W$ over $\mathbb{R}$ and "extend the scalars" to turn it into a $\mathbb{C}$-vector space.
For a concrete example, $\mathbb{R}$ is certainly an $\mathbb{R}$-vector space in an obvious way; is there a natural way to view $\mathbb{R}$ as a $\mathbb{C}$-vector space?
The point is that in a $k$-vector space, I have to know how to multiply vectors by every element of $k$. Making $k$ bigger (e.g. going from $\mathbb{R}$ to $\mathbb{C}$) means that I need to know more things.
Similarly, any complex inner product space "is" a real inner product space (or can be viewed as one by "restricting scalars"). However, the converse is false: there is no obvious way to turn a real inner product space into a complex inner product space.
Essentially the same phenomenon is on display here and here.
