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!
Best Answer
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.
Hope this helps!