Given arbitrary singular square matrix $A$ deficient in rank by $1$, and vector $v$, in numerical experiments with numerous values of the variables, I always get $v^T(A+vv^T)^{-1}v=1$. Why does $A$ cancel out like this? (And when not, aside from obvious zero situations?)
EDIT: My $A$ matrices have been symmetric positive semidefinite (so far). And the relation keeps holding when many or all elements of $v$ are tiny. (It takes around 6 orders of magnitude smaller than the A elements to break down).
Best Answer
Problem 1: Let $A$ be a $n\times n$ complex matrix with $\mathrm{rank}(A) = n - 1$. Let $u, v$ be two $n\times 1$ complex vectors. Suppose that $A + vu^\mathsf{H}$ is invertible where $(\cdot)^{\mathsf{H}}$ is the conjugate transpose. Prove that $$u^\mathsf{H}(A + vu^\mathsf{H})^{-1}v = 1.$$
Proof:
Note that $$A = A + vu^\mathsf{H} - vu^\mathsf{H} = \big(I_n - vu^\mathsf{H}(A + vu^\mathsf{H})^{-1}\big)\,(A + vu^\mathsf{H}).$$ Thus, we have $$\det A = \det \big(I_n - vu^\mathsf{H}(A + vu^\mathsf{H})^{-1}\big) \,\det (A + vu^\mathsf{H}) $$ which results in $$\det \big(I_n - vu^\mathsf{H}(A + vu^\mathsf{H})^{-1}\big) = 0.$$ Using $\det (I_n + xy^\mathsf{H}) = 1 + y^\mathsf{H}x$, we have $1 - u^\mathsf{H}(A + vu^\mathsf{H})^{-1}v = 0$.
We are done.