mattmns
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Hello, there is this question in the book:
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Consider the vector space of functions defined for t>0. Show that the following pairs of functions are linearly independent.
(a) t, 1/t
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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 a + \frac{-b}{t^2} = 0
which implies a = \frac{b}{t^2}
If we plug this into the first equation, we get \frac{b}{t^2}t + \frac{b}{t} = 0
so, \frac{2b}{t} = 0 => b = 0
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.
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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?Thanks!
---------
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 a + \frac{-b}{t^2} = 0
which implies a = \frac{b}{t^2}
If we plug this into the first equation, we get \frac{b}{t^2}t + \frac{b}{t} = 0
so, \frac{2b}{t} = 0 => b = 0
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?Thanks!
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