# Proving linear independence of two functions in a vector space

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• fatpotato
Hello,Thank you for your contribution!In summary, the conversation is about proving the linear independence of two simple functions in a vector space. The exercise is from a textbook, but the solution seems incoherent. The solution involves showing that the only solution to the linear combination of the two functions is the trivial solution, where the coefficients are both zero. However, the author's correction states that the functions are linearly independent, but the individual values of the functions are not taken into account. This is clarified by the fact that the zero here refers to the zero function, which is zero for all arguments, not the number zero.
fatpotato
TL;DR Summary
Proving linear independence of two simple functions in a vector space. Exercice is from a textbook, but solution seems incoherent.
Hello,

I am doing a vector space exercise involving functions using the free linear algebra book from Jim Hefferon (available for free at http://joshua.smcvt.edu/linearalgebra/book.pdf) and I have trouble with the author's solution for problem II.1.24 (a) of page 117, which goes like this :

Prove that each set ##\{f,g\}## is linearly independant in the vector space of all functions ## \mathbb{R}^+ \rightarrow \mathbb{R}##. In this case at point (a) of exercise with ##f(x) = x , g(x) = \frac{1}{x}##

If I understand correctly, I need to show that the only solution to the linear combination of ##f## and ##g## such that ## c_1 f + c_2 g = 0## is the trivial solution, where ## c_1 = c_2 = 0##.

By choosing ##c_1 = 1## and ##c_2 = -1##, I get :

$$c_1 f + c_2 g = 0 \iff c_1f = -c_2g \iff f = g \iff x = \frac{1}{x}$$

Now, there is indeed a value of ##x## such that this equation is satisfied for non-zero coefficients, with ##x=1##, thus implying that, for at least one value of ##x##, there is a solution to the previous linear combination, so the functions are linearly dependant.

However, the author clearly says that these functions are linearly independant (see correction at http://joshua.smcvt.edu/linearalgebra/jhanswer.pdf#ans.Two.II.1.24). What am I doing wrong? Should I only take the two functions in their most general sense, without evaluating them for a given ##x##?

Thank you.

Edit : spelling

Last edited by a moderator:
The linear dependence has to hold for all x, not just some.

WWGD and fatpotato
Hello,

Thank you, I was not sure whether a single occurrence would prove linear dependance. I will keep in mind that it has to hold for all x.

Best regards

fatpotato said:
Summary:: Proving linear independance of two simple functions in a vector space. Exercice is from a textbook, but solution seems incoherent.

Hello,

I am doing a vector space exercise involving functions using the free linear algebra book from Jim Hefferon (available for free at http://joshua.smcvt.edu/linearalgebra/book.pdf) and I have trouble with the author's solution for problem II.1.24 (a) of page 117, which goes like this :

Prove that each set ##\{f,g\}## is linearly independant in the vector space of all functions ## \mathbb{R}^+ \rightarrow \mathbb{R}##. In this case at point (a) of exercise with ##f(x) = x , g(x) = \frac{1}{x}##

If I understand correctly, I need to show that the only solution to the linear combination of ##f## and ##g## such that ## c_1 f + c_2 g = 0## is the trivial solution, where ## c_1 = c_2 = 0##.

By choosing ##c_1 = 1## and ##c_2 = -1##, I get :

$$c_1 f + c_2 g = 0 \iff c_1f = -c_2g \iff f = g \iff x = \frac{1}{x}$$

Now, there is indeed a value of ##x## such that this equation is satisfied for non-zero coefficients, with ##x=1##, thus implying that, for at least one value of ##x##, there is a solution to the previous linear combination, so the functions are linearly dependant.

However, the author clearly says that these functions are linearly independant (see correction at http://joshua.smcvt.edu/linearalgebra/jhanswer.pdf#ans.Two.II.1.24). What am I doing wrong? Should I only take the two functions in their most general sense, without evaluating them for a given ##x##?

Thank you.

Edit : spelling
Like @martinbn wrote, the 0 here is the zero _function_ , which is 0 for all arguments and not the _ number_ 0.

fatpotato
WWGD said:
which is 0 for all arguments
This is what clicked for me. As a shortcut, I assumed it had to be zero, and not zero for all arguments.

WWGD
fatpotato said:
This is what clicked for me. As a shortcut, I assumed it had to be zero, and not zero for all arguments.

With your definition of linear independence, how could any two functions ever be linearly indepedent? Let ##f(x)## and ##g(x)## be functions. Take any value ##x = a##. If ##f(a) = 0## or ##g(a) = 0##, then we have ##f(a) + 0.g(a) = 0## etc. And if ##f(a), g(a) \ne 0##, then ##f(a) - \frac{f(a)}{g(a)} g(a) = 0##.

In other words, any two numbers cannot be linearly independent.

## 1. How do you define linear independence of two functions in a vector space?

Linear independence of two functions in a vector space means that one function cannot be expressed as a linear combination of the other function. This means that the two functions are not dependent on each other and can exist separately.

## 2. What is the process for proving linear independence of two functions in a vector space?

The process for proving linear independence of two functions in a vector space involves setting up a system of equations with the two functions as the variables. Then, solving the system of equations to see if there is a non-trivial solution (a solution where at least one of the variables is not equal to zero). If there is no non-trivial solution, then the two functions are linearly independent.

## 3. Can two functions with different domains and ranges be linearly independent?

Yes, two functions with different domains and ranges can still be linearly independent. The domain and range do not affect the linear independence of the functions, as long as they can still be expressed as a linear combination of each other.

## 4. Is it possible for two functions to be linearly independent but have the same graph?

Yes, it is possible for two functions to be linearly independent but have the same graph. This can happen if the functions have different coefficients or constants in their equations, but still produce the same graph.

## 5. Can more than two functions be linearly independent in a vector space?

Yes, more than two functions can be linearly independent in a vector space. The concept of linear independence can be extended to any number of functions, not just two. In general, n functions are linearly independent if none of them can be expressed as a linear combination of the others.

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