Under what changes to one specific column, is a matrix's determinant invariant? Given a fully-ranked, square matrix of columns $A =[v_1, v_2,...v_n]$. Can someone help me get some intuitive understanding of the set of vectors $u$ such that $det(A) = det( [u, v_2,...v_n] )$? Is there a way to describe this set? Is there a class of linear transformations that given $v_1$, produce those vectors $u$ that maintain the same determinant of A?
 A: Write $u=v_1+w$, so $w=u-v_1$.  Then $\det(u,v_2,\ldots,v_n)=\det(v_1,v_2,\ldots,v_n)+\det(w,v_2,\ldots,v_n)$, so the determinant is unchanged if and only if $\det(w,v_2,\ldots,v_n)=0$.  This happens if and only if $\{w,v_2,\ldots,v_n\}$ is linearly dependent.  Unless $\{v_2,\ldots,v_n\}$ is already linearly dependent, in which case all of these determinants are $0$, this is equivalent to $w$ being a linear combination of $\{v_2,v_3,\ldots,v_n\}$.
Since you assumed the original matrix is fully ranked, the answer is that the new column is the old one plus a linear combination of the others.
A: Suppose
$$
    \det([u, v_2, \dots, v_n]) = \det([v_1, \dots, v_n])
$$
Since $A = [v_1, \dots, v_n]$ is fully-ranked, we know
$$
    u = c_1 v_1 + c_2 v_2 + \dots + c_n v_n 
$$
for some constants $c_1, \dots, c_n$.  Then
\begin{align*}
    \det([u, v_2, \dots, v_n]) &= \det([c_1 v_1 + c_2 v_2 + \dots + c_n v_n,v_2,\dots, v_n]) \\
&= c_1 \det([v_1,v_2,\dots,v_n]) + c_2\det([v_2,v_2,\dots,v_n]) + \dots +c_n \det([v_n,v_2,\dots,v_n])
\end{align*}
since the determinant is multilinear.  All but the first determinant in the right-hand side above repeat one of the vectors.  Since the determinant is skew-symmetric, each of those determinants is zero.  Therefore
$$
    \det([u, v_2, \dots, v_n]) = c_1 \det([v_1,v_2,\dots,v_n])
$$
and this is satisfied as long as $c_1 =1$.  Therefore it is necessary and sufficient that $v_1-u$ be a linear combination of $v_2, \dots v_n$.
A: Let's denote by $\varphi \colon \mathbb{F}^n \rightarrow \mathbb{F}$ the map $\varphi(v) = \det([v, v_2, \dots, v_n])$. By the properties of the determinant, $\varphi$ is a linear functional and by assumption, it is non-zero as $\varphi(v_1) \neq 0$. Set $c = \varphi(v_1)$. The set you are looking for is just 
$$ S := \{ v \in \mathbb{F}^n \, | \, \varphi(v) = c \} = \phi^{-1}(c). $$
This is the solution set of a single (non-homogeneous) linear equation so it is a hyperplane in $V$. You already know one solution $v_1$ and the general solution is given by
$$ S := \{ v_1 + u \, | \, \varphi(u) = 0 \}. $$
In other words, any vector $v$ such that $\det([v,v_2,\dots,v_n]) = \det([v_1,\dots,v_n])$ is of the form $v_1 + u$ where $u$ is a vector that belongs to the $n - 1$ dimensional subspace $\operatorname{span} \{ v_2, \dots, v_n \}$.
