Question: I came across this question in one of the linear algebra textbooks:

Proof the linear independence of the following set $\{f\_i ∣ i\in\mathbb{N}\}$, such that $f\_i : \mathbb{N}\to\mathbb{Q}$ defined as follows:

For $n\in\mathbb{N}$, $$f\_i (n) = 
\begin{cases}
 -n & \text{for }n \geq i \\ 
 0 & \text{for }n < i \end{cases} $$

What would be an appropriate approach to check the independence of this infinite set? I thought long about it but was unable to come up with anything.

Any help would be appreciated.
Thanks!

Answer: 
View each function as an infinite vector  

$$v\_i = \left( \begin{array}{c} f\_i (1)\\ f\_i (2) \\ f\_i (3)\\ \vdots \end{array}\right) = \left( \begin{array}{c} 0\\0\\\vdots\\0\\-i\\ -i-1\\-i-2\\\vdots\end{array}\right)$$

in $\prod\_{n\in\mathbb{N}}\mathbb{N} $, and show that the $v\_i $ are linearly independent (this is more natural way to view it in my opinion). For example, for $i=5$, $v\_i = (0,0,0,0,-5,-6,-7,...)$

Then to check linear Independence, we check that the following is true only when all coefficients $a\_i $ are zero:
$$ \sum\_{i\geq 1} a\_i v\_i = \left( \begin{array}{c}
-a\_1 \\
-2(a\_1 + a\_2) \\
-3 (a\_1 + a\_2 + a\_3) \\
\vdots
\end{array}\right)
=
 \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ \vdots \end{array}\right)$$

But that's clear, since if we solve for the first entry we get $a\_1 = 0$. Substituting that into the next entry we get $a\_2 = 0$. Continuing to infinity, we see that $$ a\_i = 0 \qquad \mbox{ for all } i\geq 1$$
I assume that in your class $0\notin \Bbb N$ (as otherwise the claim is clearly false).

Assume $\sum\_i c\_i f\_i=0$ with almost all $c\_i=0$.
For any $n>1$, we have 
$$ 0=\sum\_i c\_if\_i(n)-\sum\_i c\_if\_i(n-1)=\sum\_i c\_i(f\_i(n)-f\_i(n-1))=-nc\_n,$$
hence $c\_n=0$. Thus $0=\sum\_i c\_if\_i(1)=c\_1f\_1(1)=-c\_1$ and also $c\_1=0$.