Show that a set of functions is linearly independent

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SUMMARY

The discussion centers on demonstrating the linear independence of the vector set a = (e-t, e-it, et, eit). The original attempt involved calculating derivatives and evaluating them at specific points, but led to confusion regarding linear dependence. A correct approach involves setting up a system of equations based on the coefficients of the exponential functions and confirming that all coefficients must equal zero. The final resolution emphasizes the necessity of constructing a 4x4 matrix for a proper analysis of the linear independence.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically linear independence.
  • Familiarity with exponential functions and their properties.
  • Basic knowledge of complex numbers and their manipulation.
  • Experience with matrix representation of linear systems.
NEXT STEPS
  • Study the properties of linear independence in vector spaces.
  • Learn how to construct and analyze 4x4 matrices for linear systems.
  • Explore the implications of complex exponentials in linear algebra.
  • Investigate the use of software tools like Maple for visualizing and solving mathematical problems.
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Mathematics students, educators, and researchers interested in linear algebra, particularly those focusing on vector spaces and complex functions.

Tala.S
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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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Tala.S said:
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 !


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 [itex]\,a=b=0\,[/itex] . Continue a little more to get that all the coefficients must be zero.
 
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.
 

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