How to prove linear independence of a function, if the function has different results?

Question

I have to prove that the following set of functions $ \{ f_i \mid i \in \ \mathbb{N}\}$ is linear independent. The function is defined as followed:

$f_i \in \textrm{Map}(\mathbb{N},\mathbb{Q})$

$f_i(n) = \left\{ \begin{array}{rl} -n & \mbox{for } n \ge i \\ 0 & \mbox{for } n < i \end{array} \right. \qquad (n \in \mathbb{N}) $

I know that I can prove linear independence for given functions by using the general criterium for linear independence.

For instance, if I have the functions $f_1(x) = cos(x)$ and $f_2(x) = sin(x)$ then I have to solve the following equation:

$a * f_1(x) + b * f_2(x) = 0$ so $$a = b = 0$$.

But how do I apply this concept to the given set of functions?

Answer 1

I think I would not use the general criterium explicitly for the proof, but rather use the fact that your functions are 0 until a given point which is different for all functions, and then have the same value. But using this as the only argument probably implies some strong undertanding of what independance means.

So if you prefer, you can use the fact that the general criterium is equivalent to the fact that if the set is not independent, then you have:

$$\exists i_0, \exists (a_1, ..., a_k): f_{i_0} = \sum_{i \neq i_0} a_if_i$$

You can see that this is equivalent to the general criterium by subtracting $f_{i_0}$ from both sides of the equation.

In your case, take any of your function $f_i$, if you try to build it as a linear combination of the other functions you reach a problem:

• you cannot add any function $f_j$ with $j<i$ because that would mean you have a non zero value where $f_i$ is zero
• you cannot, with any $f_j$ where $j>i$ have a non zero piece between $i$ and $i+1$.

I'm not sure this explanation is clear, though, is it ok?

Answer 2

Consider the set $\{f_i|i=1,2,\ldots,m\}$. To show the independece, we need to solve for $a_1,a_2,\ldots,a_m$ such that \begin{equation} \sum_{i=1}^ma_if_i(n) = 0. \end{equation} When $n=2$, $f_i(n)=0$, for $i\geq 2$. Thus, we get $a_1(-1)=0$. Therefore, $a_1=0$.

Repeating similar arguements for $n=3,4,\ldots,m$, we can show that $a_i=0$, for $i=1,2,\ldots,m$.

Since the choice of $m$ was arbitrary, it holds for any $m\in N$, and thus holds for $\{f_i|i\in N\}$