Proving linear independence of two functions in a vector space

[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

 

[martinbn]

The linear dependence has to hold for all x, not just some.

 

[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

 

[WWGD]

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]

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.

Thank you for your contribution!

 

[PeroK]

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

Thank you for your contribution!

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.