Help with a proof about inner product space

Hi guys.

I need help with the following:

Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{c}$.

Let $\displaystyle B=\left \{ v_1, v_2,...,v_n \right \}$ be a base for $\displaystyle V$.

a. let $\displaystyle v\in V$ such that for every $\displaystyle i$: $\displaystyle <v_i,v>=0$. prove that $\displaystyle v=0$.

b. let $\displaystyle u,w\in V$ such that for every $\displaystyle i$: $\displaystyle <v_i,u>=<v_i,w>$. prove that $\displaystyle u=w$.

this is what I tried so far:

a.

let $\displaystyle v_j\in B$. since $\displaystyle B$ is base for $\displaystyle V$, then:

$\displaystyle <v_j,v>=<v_j,\sum_{i=1}^{n}\alpha_iv_i>=\overline{ <\sum_{i=1}^{n}\alpha_iv_i,v_j>}=\sum_{i=1}^{n} \alpha_ i \overline{<v_i,v_j>}=...$

not sure what's next, though. what am I missing here?

Am I even in the right direction?

Thanks in advanced!

Re: Help with a proof about inner product space

Quote:

Originally Posted by

**Stormey** Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{c}$.

Let $\displaystyle B=\left \{ v_1, v_2,...,v_n \right \}$ be a base for $\displaystyle V$.

a. let $\displaystyle v\in V$ such that for every $\displaystyle i$: $\displaystyle <v_i,v>=0$. prove that $\displaystyle v=0$.

b. let $\displaystyle u,w\in V$ such that for every $\displaystyle i$: $\displaystyle <v_i,u>=<v_i,w>$. prove that $\displaystyle u=w$.

For part a):

Recall that $\displaystyle <v,v>=0\text{ if and only if }v=0$.

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

You finish the proof of part a.

Re: Help with a proof about inner product space

Thanks for the help, Plato.

Quote:

Originally Posted by

**Plato** For part a):

Recall that $\displaystyle <v,v>=0\text{ if and only if }v=0$.

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

You finish the proof of part a.

how come?

let's say j=2, how do I use this fact here:

$\displaystyle \alpha_1<v_2,v_1>+\alpha_2<v_2, v_2>+\alpha_3<v_2,v_3>+...+ \alpha_n <v_2,v_n>=0$

Re: Help with a proof about inner product space

Quote:

Originally Posted by

**Stormey** let's say j=2, how do I use this fact here:

$\displaystyle \alpha_1<v_2,v_1>+\alpha_2<v_2, v_2>+\alpha_3<v_2,v_3>+...+ \alpha_n <v_2,v_n>=0$

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

Re: Help with a proof about inner product space

Quote:

Originally Posted by

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

I'm sorry, I guess I'm missing something here.

I'll start over, and use some other letter to clarify my question.

$\displaystyle u$ is some vector in vector space $\displaystyle V$, such that $\displaystyle <v_i,u>=0$ for every i ($\displaystyle v_i\in B$, and $\displaystyle B$ 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 $\displaystyle u$ (which is some vector in $\displaystyle V$) with each of the basis vectors is 0, then u is 0"?

thanks.

Re: Help with a proof about inner product space

Quote:

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.

$\displaystyle u$ is some vector in vector space $\displaystyle V$, such that $\displaystyle <v_i,u>=0$ for every i ($\displaystyle v_i\in B$, and $\displaystyle B$ is base).

That changes nothing of what I have posted above.

$\displaystyle \{v_1,v_2,\cdots,v_n\}$ **is a basis** and the vector $\displaystyle u$ has the property that $\displaystyle \forall j<u,v_j>=0~.$

Using the basis $\displaystyle u = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}} $,

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

Re: Help with a proof about inner product space

Thank you!

now it is clear. :)