Linear Independence of two Functions

by mattmns
Tags: functions, independence, linear
mattmns is offline
Feb19-06, 06:19 PM
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Hello, there is this question in the book:
Consider the vector space of functions defined for t>0. Show that the following pairs of functions are linearly independent.
(a) t, 1/t
So if they are linearly independent then there are a,b in R, such that

at + b/t = 0

So if we differentiate both sides with respect to t we get [tex]a + \frac{-b}{t^2} = 0[/tex]

which implies [tex]a = \frac{b}{t^2}[/tex]

If we plug this into the first equation, we get [tex] \frac{b}{t^2}t + \frac{b}{t} = 0[/tex]

so, [tex]\frac{2b}{t} = 0 => b = 0[/tex]

If we plug this back to the first equation it follows that a = 0. So a = 0 and b = 0, and therefore the two functions, t and 1/t, are linearly independent.

Is that sufficient?

Also, what if I just said something like let t = 1, you then get the equation
a + b = 0
then let t = 2, and you get the equation
2a + b/2 = 0
And if you solve these equations simultaneously, it follows that a = 0 and b = 0 are the only solutions. Would that be sufficient, given that we know a = 0 and b = 0 are always solutions?

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HallsofIvy is online now
Feb19-06, 06:40 PM
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Yes, that's perfectly valid. Remember that saying at+ b/t= 0 means it is true for all t. The simplest thing to do is just choose two values for t so you have two equations to solve for a and b. Taking t= 1 tells us that a+ b= 1. Taking t= 2 gives a+ b/2= 0. Subtracting that from a+ b= 0, we have b/2= 0 so b=0. Putting b= 0 int a+ b= 0 gives a= 0 so the functions are linearly independent.
mattmns is offline
Feb19-06, 06:46 PM
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Awesome, thanks!

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