Proving inner product with linear functional

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Question: Let $V$ be a vector space over $\Bbb C$ and let $\mu :V \times V \to \Bbb C$ be a function satisfying the following conditions:

  • For each $w \in V$, the function $v \mapsto \mu (v,w)$ from $V$ to $\Bbb C$ is a linear functional
  • If $v,w \in V$ then $\mu (v,w) = \overline{ \mu(w,v)}$
  • If $v \in V$ satisfies $\mu(v,w) = 0$ for all $w\in V$, then $v=0_v$.

Is $\mu$ and inner product on $V$?


I know the distributive and commutative properties for a linear functional and I know that there are 3 (or more depending how written) conditions to satisfy for an inner product. These conditions seem very similar to those listed in the question.

I think my issue lies in not knowing how to put all of these conditions together. Any direction would be greatly appreciated.

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The second condition contains a misprint... Anyway, the third condition tells you that the map has a trivial kernel. However, an inner product must be positive-definite, which may fail even if the kernel is reduced to the null vector. For instance, you may have a negative-definite bilinear product.