Let $f_1,\ldots,f_n$ be linearly independent linear functionals on a vector space $X$. Show that there exist $x_1,\ldots,x_n\in X,$ such that the $n\times n$ matrix $\big(f_i(x_j) \big)_{i,j=1}^n$ is non-singular.


Let $F:X\to\mathbb R^n$ defined as follows $$ F(x)=\big(f_1(x),\ldots,f_n(x)\big). $$ Clearly $F$ is a linear transformation and its range $Y$ is linear subspace of $\mathbb R^n$.

We shall show that the range is the whole $\mathbb R^n$. If it is not, then we can find a vector $v\in\mathbb R^n$, such that $v=(v_1,\ldots,v_n)\perp Y$, i.e. $$ v\cdot y=0, \quad \text{for all}\,\,\,y\in Y. $$ But this means that for every $x\in X$: $$ 0=v\cdot F(x)=v\cdot \big(f_1(x),\ldots,f_n(x)\big)=v_1f_1(x)+\cdots+v_nf_n(x), $$ which means that the $f_1,\ldots,f_n$ are linearly dependent, a contradiction. Hence $F$ is onto. Let $x_1,\ldots,x_n$, such that $F(x_j)=e_j$, where $e_1,\ldots,e_n$ is the standard basis of $\mathbb R^n$. Then the matrix $$ \big(f_i(x_j)\big)_{i,j=1}^n, $$ is the unit matrix, and its determinant is equal to $1$.

Note. The proof is valid even in the case in which $\mathbb R^n$ is replaced by $\mathbb C^n$.

  • $\begingroup$ good proof…..simple yet tricky $\endgroup$ – Gaurav Feb 22 '14 at 22:47
  • $\begingroup$ Is it possible to not use any inner product on $\mathbb{R}^n$? It seems to me that it's bad style to introduce an arbitrary inner product. $\endgroup$ – CrabMan Jul 7 '20 at 13:45

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