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.
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 $\displaystyle n$).
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.
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 $\displaystyle \{e_j : j>3 \}$