Under what conditions, the rank of a matrix reduces, if it is subtracted by a vector

Question

I have a matrix $A \in \mathbb{R}^{n \times n}$ such that $\text{rank}(A)=n$. Let $B$ be a matrix defined as $B=A-ue^T$ where $u \in \mathbb{R}^n$ and $e=(1,1,\ldots,1)^T$ i.e. each column of $A$ is subtracted by vector $u$. As shown in question rank reduction by subtraction we can say that $\text{rank}(B) \geq n-1$.

My question is, can we show that rank$(B)=n-1$, if and only if vector $u$ is lying in some lower dimensional affine subspace $S \subseteq \mathbb{R}^{k}$ where $k < n$ created by some particular (not all) vectors of $A$.

Answer 1

The condition corresponds to an $n-1$ dimensional affine subspace. As $A$ is invertible, write $Q=A^{-1}$ and then e.g. $$ B = (I-u e^T Q) A$$ The rank of $B$ equals $n-1$ precisely when non-invertible, i.e. when $$ \det B = \det (I - u e^T Q) \det A = (1 - (e^T Q u)) \det A=0$$ (where I used that for a rank 1 operator $T$ we have: $\det(I-T) = 1 - {\rm tr\;} T $. This defines the wanted affine subspace: $$e^T Q u = 1.$$

Edit: Writing $u=A \lambda$ with $\lambda\in {\Bbb R}^n$ the condition reads: $e^T Q A \lambda = e^T \lambda = 1$ or simply $\lambda_1+\cdots + \lambda_n=1$ which seemingly provides a positive answer to your question (with the right interpretation).

Answer 2

I am not sure if my answer will be that one you are searching for, but I think the answer is no. Indeed let $A = (a_1,\cdots,a_n)$ and $u = a_1 - a_2$. For every $v = (v_1,\cdots,v_n)\in \mathbb{R}^n$, $B.v = \sum_{i=1}^{n} v_ia_i - \left(\sum_{i=1}^{n} v_i\right)u = \sum_{i=3}^{n}v_ia_i + \left(v_1 +\sum_{i=1}^{n} v_i\right)a_1 + \left(v_2 - \sum_{i=1}^{n} v_i\right)a_2 = 0 \Leftrightarrow \left\{\begin{array}{cc}v_i = 0, & \forall i \ge 3 \\ v_1 +\sum_{i=1}^{n} v_i = 0 & \\ v_2 - \sum_{i=1}^{n} v_i = 0\end{array}\right. \Leftrightarrow v = 0$.

This prove that $u$ could lay in $S$ created by some particular (not all) vector of $A$ but $rank(B) = n$.

To have the equivalence you must add the condition that $e^TA^{-1}u \neq 0$