Let $M$ be a $n*n$ matrix with full rank, and $g_1, \dots, g_n$ are continuous non-constant real functions with support on some open interval. Does there exist $x_0$ such that a matrix $$Q:= M-(1, \dots, 1)^\top(g_1(x_0), \dots, g_n(x_0))$$ has also full rank? Here, matrix $(1, \dots, 1)^\top(g_1(x_0), \dots, g_n(x_0))$ corresponds to n times repeating $g_1(x_0)$ in the first row, ... , $g_n(x_0)$ in the last row. I.e. a matrix $n*n$ with rank not larger than 1.
In fact, $g_i$ are somehow nice, and we can assume something more about them if that helps. It seems like this should hold for "typical" $x_0$ but I am struggling with finding at least one (I am interested only in existence of $x_0$). Some ideas?
It is helpful to note that $Q$ will fail to be full rank (i.e. invertible) if and only if $$ (g_1(x_0),\dots,g_n(x_0))M^{-1}(1,\dots,1)^\top = 1. $$ With that, we see that a satisfactory $x_0$ fails to exists if (and only if) $(g_1(x),\dots,g_n(x))$ lies within the plane $\{x : xM^{-1}(1,\dots,1)^\top = 1\}$. For instance, take any function $g_1(x)$, $g_2(x) = 1 - g_1(x)$, and $$ M = \pmatrix{1&0\\0&1}. $$ For this example, there exists no $x_0$ such that $Q$ has full rank.
Note: If the functions $g_1,\dots,g_n$ are polynomials and there exists at least one $x_0$ for which $Q$ is not full rank, then it must hold that $Q$ will be full rank for all but finitely many values of $x_0$.
If $g_1,\dots,g_n$ are holomorphic and there exists at least one $x_0$ for which $Q$ is not full rank, then it must hold that $Q$ will be full rank for all but countably many values of $x_0$.