# Linear span of functions

Maybe this is something basic but I am not familiar with the term "linear span" and in one question it mentions

the linear span of the functions $f_n(t) = e^{nt}$, $n = 0,1,2,...$ $t \in [a,b]$

What I understood is linear span is the set of linear combination of all elements but I am still unable to understand what that set is.

Thank you very much.

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Well, as you said, $f\in \operatorname{span} \{ e^{n t},\ n\in \mathbb{N}\}$ iff there exist $N\in \mathbb{N}$ and $N$ real numbers $\alpha_0, \ldots , \alpha_N$ such that: $$f(t)=\sum_{n=0}^N \alpha_n\ e^{nt}$$ (this by the very definition of linear combination), therefore: $$\operatorname{span} \{ e^{n t},\ n\in \mathbb{N}\} =\Bigg\{ \sum_{n=0}^N \alpha_n\ e^{nt},\ N\in \mathbb{N},\ \alpha_0, \ldots , \alpha_N \in \mathbb{R}\Bigg\} \; .$$

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I think there is a problem with $\sum_{n=0}^N \alpha_n\ e^{nt}$. It looks like $n$ in $e^{nt}$ cannot be greater than $N$. But it can, right? – marvinthemartian Jan 5 '12 at 18:59
@marvinthemartian note, some of the $\alpha_n$ can be 0. $N$, of course, is arbitrary. – David Mitra Jan 5 '12 at 22:48
@DavidMitra Ooooo.. okey. I got it now :) – marvinthemartian Jan 6 '12 at 9:24

The linear span of a set $V$ consists of all vectors $w$ of the form $$w=c_1 v_1+c_2v_2+\cdots+c_n v_n$$ where the $c_i$ are scalars, the $v_i$ are vectors in $V$ and $n$ is a positive integer.

So, in your case the linear span of the functions $\{f_n\}$ is the set of all finite linear combinations of the $f_n$. An element of the linear span has the form: $$\tag{1}f(t)=c_1 e^{n_1t}+c_2 e^{n_2t}+\cdots+ c_m e^{n_m t},$$ where $m$ is a positive integer, the $n_i$ are non-negative integers, and the $c_i$ are scalars.

In particular, $$f(t)=e^t-2e^{4t}+237 e^{32t}$$ is in the linear span, as is $$f(t)=3+e^t$$ $(e^{0t}=1$).

The linear span is the set of all functions of the form in (1).

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+1 for clarity. But there's a typo: your $c_ne^{c_nt}$ should be something like $c_ke^{n_kt}$. – Clive Newstead Jan 5 '12 at 15:58
@Clive Newstead Arg, thanks :) – David Mitra Jan 5 '12 at 16:00