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consider a set of vectors $H = \{e^{a_1x}, e^{a_2x}, ..., e^{a_nx}\}$ where $a_i \in \mathbb R$ and $i= 1,2,...,n$ are distinct

how can I prove that $V= \text{span}~ H$ is a linear vector space?

I know I need to prove closure under addition, scalar multiplication, and the zero vector

Could I do this by say:

$f(x) = m_1* e^{a_1x} + m_2* e^{a_2x} + ... + m_n * e^{a_nx}$

$g(x) = l_1* e^{a_1x} + l_2* e^{a_2x} + ... + l_n * e^{a_nx}$

for some $m_i$ and $l_i, i= 1,2,...,n$

and then try $f(x) + g(x)$?

and for scalar try: $r*f(x) = r (m_1* e^{a_1x} + m_2* e^{a_2x} + ... + mn * e^{a_nx})$

I am not sure what to do about the $0$ vector

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Welcome to MSE! It helps to format using MathJax. Please make sure I edited like you want. Regards –  Amzoti Feb 12 '13 at 16:42
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You can drop the assumption on the zero vector, this is for free. –  Damien L Feb 12 '13 at 16:58

3 Answers 3

If you're taking $V = \text{span}~ H$, then by definition, $V$ is the closure under addition and scalar multiplication of $H$, including the zero vector, so $V$ is a linear vector space.

Maybe you want to ask what this linear vector space looks like?

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By definition of linear span, $span(H)$ is a vector space. The zero vector being the trivial combination (all coefficients zero).

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Because all $a_i$ are distinct we can order them without loss of generality. So let them be ordered like this: $a_1<a_2<...<a_n$. Now suppose that $e^{a_jx}$ is a linear combination of $e^{a_1x},e^{a_2x},...,e^{a_{j-1}x},e^{a_{j+1}x},...,e^{a_nx}$, so that $e^{a_jx}=\sum_{i=1}^{n}b_ie^{a_ix},(i\neq j)$. (I do not know how to write that the sum excludes the case $i=j$ so I wrote it in the brackets). Now we divide this expression by $e^{a_jx}$ to get $1=\sum_{i=1}^{n}b_ie^{(a_i-a_j)x}$ and now let $x\to\infty$. When you go to the limit the terms $e^{(a_i-a_j)x}$ go to $0$ if $(a_i-a_j)<0$ and to $\infty$ if $(a_i-a_j)>0$ so the right hand side is not equal to $1$ so the vectors in the set are linearly independent and because of that they span a vector space.

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