# Thread: Orthogonal Complement

1. ## 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.

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

3. 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.

4. For example you have $\{e1, e2, e3\}$ and you want to express a sequence $(x_j)_{j \geq 1}$ in $l^2$, then what you are missing is the terms $(x_i)_{i>3}$, so that $Y^{\perp}$ should be all sequences of the form $(y_i)_{i>3}$, and a basis for $Y^{\perp}$ should be $\{e_j : j>3 \}$