# Orthogonal vectors

• May 17th 2011, 07:04 AM
alexmahone
Orthogonal vectors
Find a vector $\displaystyle (w_1,w_2,w_3,...)$ that is orthogonal to $\displaystyle v=(1,\frac{1}{2},\frac{1}{4},...)$. Compute its length $\displaystyle ||w||$.
• May 17th 2011, 07:26 AM
tonio
Quote:

Originally Posted by alexmahone
Find a vector $\displaystyle (w_1,w_2,w_3,...)$ that is orthogonal to $\displaystyle v=(1,\frac{1}{2},\frac{1}{4},...)$. Compute its length $\displaystyle ||w||$.

Orthogonal...in what vector space and with respect to what inner product??(Headbang)(Punch)(Punch)

Tonio
• May 17th 2011, 07:29 AM
alexmahone
Quote:

Originally Posted by tonio
Orthogonal...in what vector space and with respect to what inner product??(Headbang)(Punch)(Punch)

Tonio

Hilbert space with respect to dot product.
• May 17th 2011, 07:46 AM
tonio
Quote:

Originally Posted by alexmahone
Hilbert space with respect to dot product.

There are infinite Hilbert spaces ( to add "with inner product" is futile: ANY Hilbert space is a vector space with an inner product). I can think of at least

two different Hilbert spaces with their inner products to which your vector belongs, so again: what vector space with what iiner product?

Tonio
• May 17th 2011, 07:51 AM
alexmahone
Quote:

Originally Posted by tonio
There are infinite Hilbert spaces ( to add "with inner product" is futile: ANY Hilbert space is a vector space with an inner product). I can think of at least

two different Hilbert spaces with their inner products to which your vector belongs, so again: what vector space with what iiner product?

Tonio

I think the question merely requires us to find $\displaystyle w_1,w_2,w_3,...$ such that $\displaystyle w_1+\frac{w_2}{2}+\frac{w_3}{4}+...=0$.
• May 17th 2011, 08:13 AM
tonio
Quote:

Originally Posted by alexmahone
I think the question merely requires us to find $\displaystyle w_1,w_2,w_3,...$ such that $\displaystyle w_1+\frac{w_2}{2}+\frac{w_3}{4}+...=0$.

That's what I also thought, but then you already know what to look for...(Nerd)(Giggle)...

Idea: $\displaystyle \sum\limits^\infty_{n=0}\frac{1}{4^n}=\frac{4}{3} \, , \, \, \,\sum\limits^\infty_{n=1}\frac{1}{2^{2n-1}}=2\sum\limits^\infty_{n=1}\frac{1}{4^n}=2\cdot \frac{1}{3}=\frac{2}{3}$

Tonio
• May 17th 2011, 08:26 AM
alexmahone
Quote:

Originally Posted by tonio
That's what I also thought, but then you already know what to look for...(Nerd)(Giggle)...

Idea: $\displaystyle \sum\limits^\infty_{n=0}\frac{1}{4^n}=\frac{4}{3} \, , \, \, \,\sum\limits^\infty_{n=1}\frac{1}{2^{2n-1}}=2\sum\limits^\infty_{n=1}\frac{1}{4^n}=2\cdot \frac{1}{3}=\frac{2}{3}$

Tonio

• May 17th 2011, 11:54 AM
Opalg
Quote:

Originally Posted by alexmahone
I think the question merely requires us to find $\displaystyle w_1,w_2,w_3,...$ such that $\displaystyle w_1+\frac{w_2}{2}+\frac{w_3}{4}+...=0$.

You also need to ensure that $\displaystyle \sum|w_n|^2<\infty$, and in fact you need to be able to compute that sum in order to do the second part of the question.

Big hint: Can you find a solution in which only the first two coordinates of w are nonzero?
• May 17th 2011, 12:27 PM
alexmahone
Quote:

Originally Posted by Opalg
You also need to ensure that $\displaystyle \sum|w_n|^2<\infty$, and in fact you need to be able to compute that sum in order to do the second part of the question.

Big hint: Can you find a solution in which only the first two coordinates of w are nonzero?

Thanks. $\displaystyle w=(1,-2,0,...)$ and its length is $\displaystyle \sqrt{5}$.