# orthonormal basis

• April 6th 2010, 02:26 PM
surjective
orthonormal basis
Hey,

I would like to show that the sequence $\{ \delta_{k} \}_{k=1}^{\infty}=(0,0,0,\ldots,1,0,0,\ldots)$ (the k'th entry is 1) is an orthonormal basis for the space

$l^{2}(\mathbb{N})=\Big\{\{x_{k}\}_{k=1}^{\infty} \vert x_{k}\in \mathbb{C} \forall k\in \mathbb{N}, \sum_{k \in \mathbb{N}}|x_{k}|^{2} < \infty \Big\}$

I know that to show this I must:

1) Show that $\{ \delta_{k} \}_{k=1}^{\infty}$ is a basis
for $l^{2}(\mathbb{N})$
2) Show that $<\delta_{k},\delta_{k}>=1,\forall k$
3) Show that $<\delta_{k},\delta_{k}>=0, k\neq j$

My questions is only concerning point 1): I have already shown that $\{ \delta_{k} \}_{k=1}^{\infty}=(0,0,0,\ldots,1,0,0,\ldots)$ is a basis for $l^{1}(\mathbb{N})$. Can the sequence be a basis for both spaces? If yes, would it be acceptable to use the method used in the case of $l^{1}(\mathbb{N})$?

Thanks.
• April 6th 2010, 02:40 PM
Drexel28
Quote:

Originally Posted by surjective
Hey,

I would like to show that the sequence $\{ \delta_{k} \}_{k=1}^{\infty}=(0,0,0,\ldots,1,0,0,\ldots)$ (the k'th entry is 1) is an orthonormal basis for the space

$l^{2}(\mathbb{N})=\Big\{\{x_{k}\}_{k=1}^{\infty} \vert x_{k}\in \mathbb{C} \forall k\in \mathbb{N}, \sum_{k \in \mathbb{N}}|x_{k}|^{2} < \infty \Big\}$

I know that to show this I must:

1) Show that $\{ \delta_{k} \}_{k=1}^{\infty}$ is a basis
for $l^{2}(\mathbb{N})$
2) Show that $<\delta_{k},\delta_{k}>=1,\forall k$
3) Show that $<\delta_{k},\delta_{k}>=0, k\neq j$

My questions is only concerning point 1): I have already shown that $\{ \delta_{k} \}_{k=1}^{\infty}=(0,0,0,\ldots,1,0,0,\ldots)$ is a basis for $l^{1}(\mathbb{N})$. Can the sequence be a basis for both spaces? If yes, would it be acceptable to use the method used in the case of $l^{1}(\mathbb{N})$?

Thanks.

What is the inner product on the space?
• April 6th 2010, 03:06 PM
surjective
orthonormal basis
Well, the inner product

$< \{ x_{k}\}_{k=1}^{\infty}, \{ y_{k}\}_{k=1}^{\infty} >= \sum_{k \in \mathbb{N}}x_{k}\overline{y_{k}}$

is defined on $l^{2}(\mathbb{N})$ and makes it a hilbert-space. Right?
• April 6th 2010, 05:01 PM
Focus
Quote:

Originally Posted by surjective
My questions is only concerning point 1): I have already shown that $\{ \delta_{k} \}_{k=1}^{\infty}=(0,0,0,\ldots,1,0,0,\ldots)$ is a basis for $l^{1}(\mathbb{N})$. Can the sequence be a basis for both spaces? If yes, would it be acceptable to use the method used in the case of $l^{1}(\mathbb{N})$?

Quote:

Originally Posted by surjective
Well, the inner product

$< \{ x_{k}\}_{k=1}^{\infty}, \{ y_{k}\}_{k=1}^{\infty} >= \sum_{k \in \mathbb{N}}x_{k}\overline{y_{k}}$

is defined on $l^{2}(\mathbb{N})$ and makes it a hilbert-space. Right?

The fact that it is a basis does not depend on the norm of the space. Think about it, what part of $x=\sum \lambda_i x_i$ involves a norm?