proof - if in inner product space $(\alpha , \beta )= 0$ for all $\beta \in V$ then show that $\alpha = 0$

Question

Here is my approach if this holds, $( \alpha , \alpha ) = 0 $ then $\alpha = 0$. But there can be another vector for $\alpha$ so that above equation holds. I mean in mapping $$ V\times V \longrightarrow F,$$ 2 points in domain may point to 0 in F. One such point is the O vector in V. OR is that there is 1-1 mapping between $$V\times V \longrightarrow F$$ (ensures that 1 point in F has 1 pre-image in domain). Pls clarify

Answer 1

Just take $\beta=\alpha$ and use one of the axioms of inner products.

Answer 2

In this exercise $\alpha$ is fixed. So you are considering the map $$ \beta \mapsto \langle \alpha, \beta \rangle $$ You should think of $\alpha$ as a linear map rather than a vector (so as an element in the dual space). If a linear map is zero everywhere than it is zero as a point in the dual space and the vector that represents it is zero (simply follow your explanation).

If you consider the scalar product, then you have a bilinear map $$ \langle \cdot, \cdot \rangle : V \times V \to F. $$

Of course this map has many zeros (any two orthogonal vectors), but it is not what you are looking for.

Also the map $$ \beta \mapsto \langle \alpha, \beta \rangle $$

has many zeros, namely the orthogonal complement of $\alpha.$

The only map that does not have many zeros, i.e. is one-to-one is the map that associates any vector to a function (namely the linear map above). Your calculation shows that this map in injective.