Show this set of functions is linearly independent

[Ryker]

Show this set of functions is linearly independent (e^(-x), x, and e^(2x))

Homework Statement

f_{1}(x) = e^{-x}, f_{2}(x) = x, f_{3}(x) = e^{2x}

Homework Equations

Theorems and lemmas, which state that if vectors are in echelon form, they are linearly independent, and also that they are such if we can find a corresponding matrix, written in echelon form, where the number of rows is the same as the number of original vectors.

The Attempt at a Solution

I don't really know how exactly to approach this. I guess all of the above functions can be written down in the form akin to a polynomial, such that

f_{i}(x) = a_{1}e^{-x} + a_{2}x + a_{3}e^{2x}

I then put that in matrix form and I basically got

\left(\begin{array}{ccc}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> <br /> \end{array}\right),

which is a matrix in echelon form that would confim linear independence. Am I even on the right track here? Or are e^{-x} and e^{2x} covered by the same basis vector?

Thanks in advance.

 

[Mark44]

Homework Statement

f_{1}(x) = e^{-x}, f_{2}(x) = x, f_{3}(x) = e^{2x}

Homework Equations

Theorems and lemmas, which state that if vectors are in echelon form, they are linearly independent, and also that they are such if we can find a corresponding matrix, written in echelon form, where the number of rows is the same as the number of original vectors.

The Attempt at a Solution

I don't really know how exactly to approach this. I guess all of the above functions can be written down in the form akin to a polynomial, such that

f_{i}(x) = a_{1}e^{-x} + a_{2}x + a_{3}e^{2x}

I then put that in matrix form and I basically got

\left(\begin{array}{ccc}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \\<br /> <br /> \end{array}\right),

which is a matrix in echelon form that would confim linear independence.

How so?

Am I even on the right track here?

No.

Or are e^{-x} and e^{2x} covered by the same basis vector?

No. This is equivalent to saying that they are linearly dependent, which would imply that one of them is a multiple of the other, which isn't the case.

What you need to do is to show that the following equation has one and only one solution in the constants a₁, a₂, and a₃.
f(x) = a_{1}e^{-x} + a_{2}x + a_{3}e^{2x}

(I removed the subscript on f, since there really is no need for it.)

Since the equation above is identically true for all x, you can get another equation by differentiating both sides. Then you'll two equations in three unknowns. Can you think of something you can do to get another equation so that you'll have three equations in three unknowns?

 

[Ryker]

What you need to do is to show that the following equation has one and only one solution in the constants a₁, a₂, and a₃.
f(x) = a_{1}e^{-x} + a_{2}x + a_{3}e^{2x}

Since the equation above is identically true for all x, you can get another equation by differentiating both sides. Then you'll two equations in three unknowns. Can you think of something you can do to get another equation so that you'll have three equations in three unknowns?

Should I take a second derivative? I thought log might be good, as well, but since then on the right hand side I have three elements, I'd need to take a log of their sum, which doesn't get me anywhere. So what do I do here?

But the second equation would then be

df(x) = -a_{1}e^{-x} + a_{2} + 2a_{3}e^{2x}

Is that correct?

 

[Mark44]

Almost - it would be
f &#039;(x) = -a_{1}e^{-x} + a_{2} + 2a_{3}e^{2x}

How can you get a third equation?

 

[Ryker]

I don't know, should I take a second derivative, so that I eliminate a₂? I guess then I'd have:

<br /> f&#039;&#039;(x) = a_{1}e^{-x} + 4a_{3}e^{2x}.<br />

Then to test linear dependence we set f(x) to zero, and thus f'(x) and f''(x) would also be zero, I guess. Then we'd have

0 = a_{1}e^{-x} + 4a_{3}e^{2x} \Rightarrow a_{1}e^{-x} = -4a_{3}e^{2x}.

Then I'd substitute that into the second equation to get a_{2} = -6a_{3}e^{2x}.
Repeating the procedure and substituting into the first one, I get a_{3}(-3e^{2x} + 6e^{2x}x) = 0.

Since (-3e^{2x} + 6e^{2x}x) \neq 0, a_{3} = 0, and therefore by the same logic a_₁ and a_₂ are zero.

I don't know how correct this is, I'm trying my best, but the logic still seems to be eluding me.

 

[Mark44]

Yes, that's it. Now you have shown that the only solution to the equation a_{1}e^{-x} + a_{2}x + a_{3}e^{2x} = 0 is a₁ = a₂ = a₃ = 0, hence the three functions are linearly independent.

 

[Ryker]

Awesome, thanks a lot! Although I still feel there should've been another way to solving this, somehow involving matrices explicitly (this is a problem from the chapter on row equivalence of matrices). Do you think there is? I'm namely doing problems out of Curtis's Linear Algebra, and, well, he does things kind of differently I think. I'm having a lot of trouble understanding what he's trying to say and the book just doesn't sit that well with me. Still, I have no idea whether it's just me not understanding stuff or is the book really weird in a way. Do you have any experience with this book perhaps?

 

[Mark44]

I've never heard of the Curtis text.

You can represent your three equations this way:
\left(\begin{array} {c c c} a_1 &amp; a_2 &amp; a_3 \\-a_1 &amp; a_2 &amp; 2a_3 \\ a_1 &amp; 0 &amp; 4a_3\end{array}\right)<br /> \left( \begin{array} {c} e^{-x} &amp; x &amp; e^{2x} \end{array}\right) = <br /> \left( \begin{array} {c} 0 &amp; 0 &amp; 0 \end{array}\right)

If you row reduce the matrix on the left, you get the result that the only solution is the trivial solution, as before.

 

[Ryker]

Ah, OK, thanks again for your help. As for the book, it's https://www.amazon.com/dp/0387909923/?tag=pfamazon01-20 one. I'm doing first year Linear Algebra, so I don't know how appropriate it is, even though the professor has it as recommended reading. We don't seem to be following its curriculum that strictly, though, and the explanations in class are given in a more understandable way usually.