TITLE: Prove that the set $\displaystyle\ \bigg{\{} \frac{1}{x-c}\bigg{\}}_{\displaystyle\ c \in \mathbb{R}\setminus[0,1]}$ is linearly independent.
QUESTION [7 upvotes]: Question: Let $V$ be the vector space of all real valued functions defined on the unit interval $[0,1]$.
Show that the set $\displaystyle\ \bigg{\{} \frac{1}{x-c}\bigg{\}}_{\ c \in \mathbb{R}\setminus[0,1]}$ is linearly independent.
Attempt:  Assume towards a contradiction that the set is linearly dependent. So, $\exists$ a finite subset $\displaystyle\bigg{\{}\frac{1}{x-c_i}\bigg{\}}_{i=1}^n$ which is linearly dependent.
Therefore, for some $(d_1,d_2,...,d_n)\neq (0,0,..,0)$ we must have
$\displaystyle f(x)=\sum_{i=1}^n\frac{d_i}{x-c_i}=\frac{d_1(x-c_2)...(x-c_n)+d_2(x-c_1)(x-c_3)...(x-c_n)+...+d_n(x-c_1)...(x-c_{n-1})}{(x-c_1)...(x-c_n)}=0$ for all $x\in[0,1]$.
Now, the denominator is $\neq 0$ $\forall x\in [0,1]$.
So, $f(x)=0$ only when $d_1(x-c_2)...(x-c_n)+d_2(x-c_1)(x-c_3)...(x-c_n)+...+d_n(x-c_1)...(x-c_{n-1})=g(x)=0$. However, $g(x)$ is a polynomial of degree $\leq n-1$ and $g(x)=0$ for every $x \in[0,1]$, which implies that number of zeros of $g(x)$ is $>deg(g(x))$, hence $g(x)$ must be identically equal to $0 \implies (d_1,d_2,...,d_n)= (0,0,..,0) $.
Therefore, our assumption that the given set is linearly dependent is not tenable, i.e. the set is linearly independent.
Is this correct?

REPLY [2 votes]: A finite set of differentiable functions is linearly independent when its Wronskian is non-zero.
In this case the determinant leads to a vandermonde matrix \begin{align}&W(\frac{1}{x-c_1},\ldots,\frac{1}{x-c_n})
\\
&=\det\begin{pmatrix}(x-c_1)^{-1}&\cdots&(x-c_n)^{-1}\\
-(x-c_1)^{-2}&\cdots&-(x-c_n)^{-2}\\
\vdots\\
(-1)^{n-1}(n-1)!(x-c_1)^{-n}&\cdots&(-1)^{n-1}(n-1)!(x-c_n)^{-n}\end{pmatrix}\\
&=\frac{(-1)^a\prod_{k=1}^{n-1}k!}{(x-c_1)^n\cdots(x-c_n)^n}\det\begin{pmatrix} (x-c_1)^{n-1}&\cdots&(x-c_n)^{n-1}\\
\vdots\\
1&\cdots&1\end{pmatrix}\\
&=\frac{(-1)^{a'}\prod_{k=1}^{n-1}k!\prod_{i<j}(c_i-c_j)}{\prod_i(x-c_i)^n}
\end{align}
The fraction is zero iff $c_i=c_j$ for some $i\ne j$.