# Math Help - Help with a proof about inner product space

1. ## Help with a proof about inner product space

Hi guys.
I need help with the following:

Let $V$ be a vector space over $\mathbb{c}$.
Let $B=\left \{ v_1, v_2,...,v_n \right \}$ be a base for $V$.

a. let $v\in V$ such that for every $i$: $=0$. prove that $v=0$.
b. let $u,w\in V$ such that for every $i$: $=$. prove that $u=w$.

this is what I tried so far:
a.
let $v_j\in B$. since $B$ is base for $V$, then:
$==\overline{ <\sum_{i=1}^{n}\alpha_iv_i,v_j>}=\sum_{i=1}^{n} \alpha_ i \overline{}=...$

not sure what's next, though. what am I missing here?
Am I even in the right direction?

2. ## Re: Help with a proof about inner product space

Originally Posted by Stormey
Let $V$ be a vector space over $\mathbb{c}$.
Let $B=\left \{ v_1, v_2,...,v_n \right \}$ be a base for $V$.

a. let $v\in V$ such that for every $i$: $=0$. prove that $v=0$.
b. let $u,w\in V$ such that for every $i$: $=$. prove that $u=w$.

For part a):
Recall that $=0\text{ if and only if }v=0$.

Now because $v\in V$ then $v = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}}$

You finish the proof of part a.

3. ## Re: Help with a proof about inner product space

Thanks for the help, Plato.

Originally Posted by Plato
For part a):
Recall that $=0\text{ if and only if }v=0$.

Now because $v\in V$ then $v = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}}$

You finish the proof of part a.
how come?
let's say j=2, how do I use this fact here:

$\alpha_1+\alpha_2+\alpha_3+...+ \alpha_n =0$

4. ## Re: Help with a proof about inner product space

Originally Posted by Stormey
let's say j=2, how do I use this fact here:
$\alpha_1+\alpha_2+\alpha_3+...+ \alpha_n =0$

$\left( {\forall j} \right)\left[ {\left\langle {v,{v_j}} \right\rangle = 0} \right]$

5. ## Re: Help with a proof about inner product space

Originally Posted by Plato
$\left( {\forall j} \right)\left[ {\left\langle {v,{v_j}} \right\rangle = 0} \right]$
I'm sorry, I guess I'm missing something here.

I'll start over, and use some other letter to clarify my question.
$u$ is some vector in vector space $V$, such that $=0$ for every i ( $v_i\in B$, and $B$ is base).
how do I go from: "a dot product of vector with himself is 0 iff it is 0 itself" to: if the dot product of $u$ (which is some vector in $V$) with each of the basis vectors is 0, then u is 0"?

thanks.

6. ## Re: Help with a proof about inner product space

Originally Posted by Stormey
I'm sorry, I guess I'm missing something here.
I'll start over, and use some other letter to clarify my question.
$u$ is some vector in vector space $V$, such that $=0$ for every i ( $v_i\in B$, and $B$ is base).

That changes nothing of what I have posted above.
$\{v_1,v_2,\cdots,v_n\}$ is a basis and the vector $u$ has the property that $\forall j=0~.$

Using the basis $u = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}}$,
so $\left\langle {u,u} \right\rangle = \left\langle {u,\sum\limits_{k = 1}^n {{\alpha _k}{v_k}} } \right\rangle = \sum\limits_{k = 1}^n {{\alpha _k}\left\langle {u,{v_k}} \right\rangle } = 0$

7. ## Re: Help with a proof about inner product space

Thank you!
now it is clear.