# Orthogonal Complement

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• May 31st 2009, 10:44 AM
frater_cp
Orthogonal Complement
What is Y^{\perp} if Y= span{e_1,e_2, \cdots , e_n} \subset \mathbb{l}^2, where e_j=(\delta_{ij}) ?

e_1 = (1,0,0,0, \ldots)
e_2 = (0,1,0,0, \ldots) etc.
• May 31st 2009, 11:14 AM
Jose27
It depends: Is your $\displaystyle n$ a fixed integer or not? If it's not, then the orthogonal complement is $\displaystyle \{0\}$ because $\displaystyle \{e_n : n \in \mathbb{N}\}$ is dense in $\displaystyle l^2$. If, on the other hand, $\displaystyle n$ is fixed, then $\displaystyle Y$ is a finite dimensional subspace of $\displaystyle l^2$ and so is closed, therefore you have $\displaystyle Y \oplus Y^{\perp}=l^2$. (It should be clear from here who $\displaystyle Y^{\perp}$ is in this last case, since to represent every element of $\displaystyle l^2$ you need all the elements in your sequence wich are greater or equal than the given $\displaystyle n$).
• May 31st 2009, 11:51 AM
frater_cp
Hi Jose.

Thank you for your message! Very helpful man.
So in the second case assuming n is fixed at say n=3.
How would I write the Y^{\perp} in set notation?

say x = (\xi_j) \in l^2 is it then:

Y^{\perp} = \{ x = (\xi_j) \vert \sum^{j=4}^{\infty} < \infty \} ?

or should I express interms of e_n 's.
• May 31st 2009, 01:12 PM
Jose27
For example you have $\displaystyle \{e1, e2, e3\}$ and you want to express a sequence $\displaystyle (x_j)_{j \geq 1}$ in $\displaystyle l^2$, then what you are missing is the terms $\displaystyle (x_i)_{i>3}$, so that $\displaystyle Y^{\perp}$ should be all sequences of the form $\displaystyle (y_i)_{i>3}$, and a basis for $\displaystyle Y^{\perp}$ should be $\displaystyle \{e_j : j>3 \}$