Consider a linear space with inner product $V$ and vectors $x,y \in V$. Show that if $<v,y> = <v,x>$ for all $v\in V$, then $x=y$.
My attempt:
So what i basically did was to develop the equality given applying the inner product properties:
$<v,y> = <v,x>$
$<v,y> - <v,x> = 0$
$<v,y> + <v,-x> = 0$
$<v,y-x> = 0$
And then I said that because of this fact $v$ is orthogonal to $y-x$. Can I immediately conclude with this information plus the information that $<v,y> = <v,x>$ that we have $x=y$? Or do I need to prove anything else?
Thanks!
@ Granger:
If you only assume that $V$ is a vector space over some field $k$ with an inner product $< , >$ then your second statement is not true in general. This statement actually defines a special kind of an inner product on the space $V$.
For example consider $V=\mathbb{R}^2$ and define the inner product on $V$ as $<x,y>=x_1y_1$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Consider $x=(0,1)\in V$ and $y=(0,2)\in V$. Then we have $<v,x>=0=<v,y>$ for all $v\in V$ but $x\neq y$.
On the other hand if your inner product is defined by $<x,y>=x_1y_1+x_2y_2$ then the statement you made is true. This can be easily checked since we have a concrete formula for the inner product.
More generally, suppose that the given inner product be such that the following condition holds:
Let $\{e_1,...,e_n\}$ be a basis of $V$. Construct a matrix $(a_{ij})$ by defining the $ij$-th entry as $a_{ij}=<e_i,e_j>$. Then the following two statements are equivalent:
(a) $Det(a_{ij})\neq 0$
(b) If $<v,x>=<v,y>$ for all vectors $v\in V$ then $x=y$.
The proof is not difficult. One can consult a decent Linear algebra book (for example, "Linear Algebra" by Hoffman and Kunze) for these ideas. I hope that this answer helps.