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Based on the answer to this question, if we have a matrix $$M(x_1, \dots, x_n)= \begin{bmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\f_1(x_n) & \cdots & f_n(x_n) \end{bmatrix} $$ and this matrix is not full rank, then there exists a function $g:\mathbb{R}^n \rightarrow \mathbb{R}^n\setminus \{0\}$ such that $$ M(x_1, \dots, x_n)g(x_1, \dots, x_n)=0\,\,.$$ why is this the case -- why is $g$ denoted to depend on $x_1,\dots, x_n$? I thought that a square matrix not being full rank implies that its columns are linearly dependent and would this not imply that $$ M(x_1, \dots, x_n)\mathbf{v}=0$$ for some non-zero constant vector $\mathbf{v}$? Is it just that the vector depends on the values the functions operate on, or is there something else going on?

Lulu
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1 Answers1

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Consider two separate sets of points, $x_{1,1},\dots,x_{n,1}$ and $x_{1,2},\dots,x_{n,2}$. Let $M_j = M(x_{1,j},\dots,x_{n,j})$ for $j = 1,2$.

Because $M_1$ does not have full rank, there is necessarily a vector $\mathbf v_1$ for which $M_1 \mathbf v_1 = 0$. Similarly, because $M_2$ does not have full rank, there is a vector $\mathbf v_2$ for which $M_2 \mathbf v_2 = 0$. However, the fact that $M_1$ and $M_2$ fail to have full rank is not enough to deduce that there is a single vector $\mathbf v$ for which $M_1 \mathbf v = 0$ and $M_2 \mathbf v = 0$.

Likewise, the fact that $M(x_1,\dots,x_n)$ fails to have full rank for all sets tells us that for each set $x_1,\dots,x_n$, there is a vector $\mathbf v$ for which $M(x_1,\dots,x_n)\mathbf v = 0$ (i.e. there is a possible value for $g(x_1,\dots,x_n) = \mathbf v$). However, we have no reason to believe (without proof) that there exists a single vector $\mathbf v$ such that for all $x_1,\dots,x_n,$ $M(x_1,\dots,x_n)\mathbf v = 0$.

Ben Grossmann
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