Can $A + uv^T$ be non-singular if A(order n) has rank n-1

Question

Suppose $A$ is a matrix of order $n$ with rank n-1, $u$ and $v$ are $n$-vectors. Can $A + uv^T$ be nonsingular, if yes then find such $u$ and $v$.

First I wrote Rank$(A + uv^T) <=$ Rank$(A)$ + Rank$(uv^T)$

Then I wrote $n-1 <=$ Rank$(A + uv^T) <= n$ (which I am not sure of)

Then tried to take $v$ in the kernel of $A$ but to no avail. How to approach?

Answer 1

Trivially, yes: take u=(1,0,0..,0)(column vector), v=(1,0,0...,0) and A to be the matrix that has 1 everywhere on the diagonal except at (1,1) (and 0 everywhere else). Then $A + uv^T= I$.

Adding in my comment, for the (not so trivial)case of an arbitrary fixed matrix A of rank (n-1):

$A$ is of rank (n-1) => there are invertible matrices(corresponding to elementary transforms) X,Y such that $A=XI_{n−1}Y$ where by $I_{n−1}$ I denoted the matrix with 1 on the main diagonal everywhere except at position (1,1). But now choose $u=Xu, v=Y^Tv$, where u,v are the columns I defined above in my answer and then you get that $A+uv^T=X(I_{n−1}+uv^T)Y=XY $ is invertible

Answer 2

If either u or v are in the row space or column space of A respectively then A + uv’ will remain singular. So if A = ab’ with a and b vectors (A is rank one) and say v=const * b then A + uv’ will remain a rank one matrix.