# How to prove these functions are linearly independent?

Let $$f_1, ..., f_n$$ be a set of bounded functions defined as $$f_j:V \subset \mathbb{R}^n \rightarrow \mathbb{R}$$ for $$j = 1,...,n$$.

Here is the catch, if there exists a $$v_1,...,v_n \in V$$ such that the following matrix $$F$$ is invertible:

$$\begin{bmatrix} f_1(v_1)&...&f_n(v_1)\\ \vdots&\ddots&\vdots\\ f_1(v_n)&...&f_n(v_n)\\ \end{bmatrix}$$

then $$f_1, ... , f_n$$ are linearly independent.

In my own attempt I supposed the functions were actually linearly dependent.

So they would form a column which is a linear combinations of the $$n-1$$ columns of the matrix F.

Therefore, F is not invertible. (Contradicts hypothesis)

However, I feel like I’m missing the connection between $$v_1, ..., v_n$$ and the fact that $$f_1,...,f_n$$ are linearly independent.

Could you guys help me with this one? Cheers!

The missing connection is the complete definition of linear dependence of functions. A set of functions $$f_1, f_2 \dots f_n$$ are said to be linearly dependent if there exist constants $$c_1, c_2, \dots c_n$$, all not zero, such that $$\displaystyle \sum_{i = 1}^n c_i f_i(v) = 0 \$$ for all $$v \in V$$. And if there does not exist such a set of constants then the functions are said to be linearly independent.
So now consider exactly your approach. Suppose the functions are linearly dependent. This implies that there exist constants $$c_1, c_2, \dots c_n$$, all not zero, such that $$\displaystyle \sum_{i = 1}^n c_i f_i(v) = 0 \$$ for all $$v \in V$$. And in particular for $$v_1, v_2, \dots v_n$$ as given in the question. However, we know that the matrix $$F$$ is invertible and hence the only element in its null space is the zero vector. This implies the only possible way the above equation holds is when all $$c_1, c_2 \dots c_n$$ are zero, which implies the functions are linearly independent.