Inner product: $(x,z)=(y,z)\implies x=y$?

Question

We've talked about inner products in our last tutorial and couldn't really get answered the following questions:

Let $(\cdot,\cdot)$ be any inner product. If $(x,z)=(y,z)$ for all $z$ of any given inner product space, can you follow $x=y$? If so, why?

And also, does $(x,0)=0$ always hold?

Answer 1

Let me first answer the second question: yes, $(x,0)= 0$ holds. To see this note $(x,0)= (x, 0 +0)= (x,0)+(x,0)$ and the claim follows.

Conversely, if you have some $u$ such that $(x,u)=0$ for all $x$, then $u=0$. To see this just take $x=u$ (or $x= \overline{u}$ if you have a complex space) and recall that $(u,u)=0$ if and only if $u=0$ (as an inner product is positive definite).

This allows to answer you first question by noting that $(x,z)=(y,z)$ is the same as $(x-y,z)= (x,z)-(y,z)=0$.

Answer 2

$(x,0)=(x,0+0)=(x,0)+(x,0)\Rightarrow ???$

Answer 3

If $(x,z) = (y,z)$, $\forall$ $z \in I$, where $I$ some inner product space, then:

$(x,z) - (y,z) = 0$. Using linearity with respect to the first component, $(x-y,z) = 0$. Since this holds $\forall$ $z$, then in particular for $z = x - y$. Thus:

$(x-y, x-y) = 0$, which gives $x - y = 0$, because $0$ is the only vector orthogonal to itself in an inner product space.

For your second question, we have $\forall$ $x$:

$(x,0) = 0 \times (x,0) = 0$.