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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!

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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!

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  • $\begingroup$ Thanks a bunch! It helped me see it more clearly, definitely! $\endgroup$ – upStoneLock May 15 at 12:48
  • $\begingroup$ Glad I could help! :) $\endgroup$ – sudeep5221 May 15 at 13:22

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