In section 7.A of $\textit{Linear Algebra Done Right}$, Axler proved the following theorem: Suppose $V$ is a complex inner product space and $T\in\mathcal{L}(V)$. If $$\langle Tv, v \rangle = 0$$ for all $v\in V$, then $T = 0$.
Now, define $\phi:\mathcal{L}(V)\to V'$ by $$\big(\phi(T)\big)v = \langle Tv, v \rangle$$ then $\phi$ is clearly linear by the linearity of $T$ and bilinearity of the inner product. Hence, Axler's result can be phrased as $$\ker\phi = \{0\}\,.$$ In other words, $\phi$ is a $\textit{monomorphism}$ from $\mathcal{L}(V)$ to $V'$. However, this cannot be true because $$\dim\mathcal{L}(V) = (\dim V)^2 >\dim V = \dim V'$$ assuming $\dim V$ is sufficiently large. I know my reasoning must be wrong somewhere along the line, but I could not figure out what the problem is, so please illuminate me.
At first, I thought the problem might have to do with $\mathbb{C}\cong \mathbb{R}^2$, but the same problem occurs in the real case. In fact, Axler proved in the same section the following: If T is a $\textit{self-adjoint}$ operator on a $\textit{real}$ inner product space $V$ such that $$\langle Tv, v\rangle = 0$$ for all $v\in V$, then $T = 0$. With the same definition, $\phi$ is again a linear injection from $\mathcal{S}(V)$ to $V'$, where $\mathcal{S}(V)$ denotes the space of self-adjoint operators on $V$. However, $$\dim \mathcal{S}(V) = \frac{\dim V(\dim V+1)}{2} > \dim V = \dim V'$$ for large $\dim V$. Again, there must be a loophole in my argument, so please point it out for me. Thank you!
The map $\phi(T) \colon V \to \mathbb{C}$ is in general not linear because $$ \phi(T)(\lambda v) = \langle T(\lambda v), \lambda v \rangle = \langle \lambda T(v), \lambda v \rangle = |\lambda|^2 \langle T(v), v \rangle = |\lambda|^2 \phi(T)(v). $$ The addativity of $\phi(T)$ also fails for general $T$. The map $\phi$ is therefore not even well-defined.