# Show that a set of functions is linearly independent

1. Nov 10, 2012

### Tala.S

Hello everybody

I have to show that this set of vectors a = (e-t, e-it, et, eit ) is linearly independent.

My attempt :

f(x) = k1 * e-t + k2 * e-it + k3 * et + k4 * eit

f '(x) = k1 * -e-t + k2 * -ie-it + k3 * et + k4 * ieit

f ''(x) = k1 * e-t + k2 * -e-it + k3 * et + k4 * -eit

f(0) = k1 * 1 + k2 * 1 + k3 * 1+ k4 * 1 = 0

f '(0) = k1 * -1 + k2 * -i + k3 * 1+ k4 * i = 0

f ''(0) = k1 * 1 + k2 * -1 + k3 * 1+ k4 * -1 = 0

But when I plot this in maple and reduce it I get this :

- Image -

They're linearly dependent but this can't be correct. So I guess I'm doing something wrong but what ?

I would really appreciate it if someone could help me.

I HAVE SOLVED THIS !

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Last edited: Nov 10, 2012
2. Nov 10, 2012

### DonAntonio

I'm going to guess that you actually meant to write powers and that you meant dependence over the complex there...so suppose

$$ae^t+be^{-t}+ce^{it}+de^{-it}=0\,\,\,,\,\,a,b,c,d\in\Bbb C\,\,:$$

$$(1)\;\;t=0:\Longrightarrow a+b+c+d=0$$

$$(2)\;\;t=2\pi:\Longrightarrow ae^{2\pi}+be^{-2\pi}+c+d=0\Longrightarrow a\left(e^{2\pi}-1\right)+b\left(e^{-2\pi}-1\right)=0$$

$$(3)\;\;t=-2\pi:\Longrightarrow ae^{-2\pi}+be^{2\pi}+c+d=0\Longrightarrow a\left(e^{-2\pi}-1\right)+b\left(e^{2\pi}-1\right)=0$$

From (2)-(3) you already get $\,a=b=0\,$ . Continue a little more to get that all the coefficients must be zero.

3. Nov 12, 2012

### Vargo

DonAntonio's solution makes sense.

But, OP, your solution is not clear. You should be starting with a 4x4 matrix, not a 3x4 matrix. Presumably, you would also need to write up an explanation of what those matrices have to do with the original question.