Let $M$ be some square matrix. Take some column in $M$ and add a vector $\epsilon v$ to it where $\epsilon$ is sufficiently small. Replace this the column in $M$ with this new column vector, call this matrix $M_\epsilon(v)$. I'm interested in how the determinant changes under this operation. Namely, under what conditions do we have: $$det(M_\epsilon(v) ) > det(M)$$ as $\epsilon$ tends to $0$.
Any references would be appreciated.
The determinant is a linear function of each column. If $\epsilon v$ is added to the $k^\text{th}$ column of $M$ then $$ \det (M_\epsilon(v)) = \det (M) + \epsilon \det (M') $$ where $M'$ is the matrix $M$ with the $k^\text{th}$ column replaces by $v$. In particular, $$ \det (M_\epsilon(v)) > \det (M) $$ cannot hold for all (or all sufficiently small) $\epsilon$ if $\epsilon$ can take both positive and negative values.
With the additional restriction $\epsilon > 0$ the desired estimate holds exactly if $\det(M') > 0$.