linear independence

**alexandros** · July 1st 2009, 07:32 AM

CAN somebody,please write down the definition for the linear independence of the following functions?

{ $e^x,e^{2x}$ }

**Jose27** · July 1st 2009, 08:25 AM

Originally Posted by alexandros

CAN somebody,please write down the definition for the linear independence of the following functions?

{ $e^x,e^{2x}$ }

As in any vector space, the set $\{e^x, e^{2x}\}$ is linearly independent iff whenever there are scalars $a,b \in \mathbb{F}$ (where $\mathbb{F}$ is your underlying field) such that $ae^x + be^{2x} =0$ we must have $a=b=0$ .

In this particular case putting $x=0$ we have $a=b$ and so $ae^x(1+e^x)=0$ but $0 \neq e^x$ for all $x \in \mathbb{F}$ (assuming $\mathbb{F} = \mathbb{R} or \mathbb{C}$ ) so $a(1+e^x)=0$ for all $x \in \mathbb{F}$ and since $e^x$ is not constant we must have $b=a=0$ and so your set is linearly independent.

**alunw** · July 1st 2009, 08:41 AM

Your argument is not quite correct. At x=0 we have a+b=0 i.e. a=-b

**Krizalid** · July 1st 2009, 03:23 PM

Use the Wronskian and you'll find linear indepencence.

**alexandros** · July 3rd 2009, 05:02 AM

Originally Posted by Krizalid

Use the Wronskian and you'll find linear indepencence.

I am sorry i am iterested in the definition and not the proof
thanks

**hatsoff** · July 3rd 2009, 05:51 AM

Originally Posted by alexandros

I am sorry i am iterested in the definition and not the proof
thanks

Jose27 gave you the definition:

Originally Posted by Jose27

As in any vector space, the set $\{e^x, e^{2x}\}$ is linearly independent iff whenever there are scalars $a,b \in \mathbb{F}$ (where $\mathbb{F}$ is your underlying field) such that $ae^x + be^{2x} =0$ we must have $a=b=0$ .

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