Help me prove or disprove $v^T(A+vv^T)^{-1}v=1$ for singular A

Question

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).

Answer 1

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.

Answer 2

Let $A$ be an arbitrary invertible matrix, and let $\alpha = v^{\mathsf T}A^{-1}v$. Then by the Sherman-Morrison formula,

\begin{align} v^{\mathsf T}(A + vv^{\mathsf T})^{-1}v &= v^{\mathsf T}\left(A^{-1} - \frac{A^{-1}v v^{\mathsf T}A^{-1}}{1 + v^{\mathsf T}A^{-1}v}\right)v \\ &= v^{\mathsf T} A^{-1}v - \frac{v^{\mathsf T}A^{-1}v v^{\mathsf T}A^{-1}v}{1 + v^{\mathsf T}A^{-1}v} \\ &= \alpha - \frac{\alpha^2}{1+\alpha} = \frac{\alpha}{1 + \alpha}. \end{align} Now all we have to do is extend this to the non-invertible matrix $A$ by some kind of limit operation.

To do this, rewrite everything in terms of adjugate matrices. This means $\alpha$ becomes $\frac{v^{\mathsf T} \mathrm{adj}(A) v}{\det(A)}$, so that $\frac{\alpha}{1+\alpha}$ simplifies to $\frac{v^{\mathsf T} \mathrm{adj}(A) v}{\det(A) + v^{\mathsf T} \mathrm{adj}(A) v}$. We get the equation $$ \frac{v^{\mathsf T}\mathrm{adj}(A + vv^{\mathsf T})v}{\det(A + vv^{\mathsf T})} = \frac{v^{\mathsf T} \mathrm{adj}(A) v}{\det(A) + v^{\mathsf T} \mathrm{adj}(A) v} $$ which we proved above for invertible $A$. But both sides are rational functions of the entries of $A$ and $v$, and invertible matrices are dense in the set of all matrices. So the equation must actually hold for all $A$ and $v$ in all cases when the denominators do not vanish. In particular, when $\det(A) = 0$, provided that $v^{\mathsf T} \mathrm{adj}(A) v \ne 0$, the right-hand side simplifies to $1$, proving the identity we wanted.

The problematic cases $v^{\mathsf T} \mathrm{adj}(A) v = 0$ occur exactly when $A + vv^{\mathsf T}$ is not a full-rank matrix, in which case the inverse we're taking does not exist.

Answer 3

In general, over any field, if $A$ is singular but $A+uv^T$ is nonsingular, then $v^T(A+uv^T)^{-1}u=1$.

Let $(A+uv^T)x=u$. Then $Ax=(1-v^Tx)u$. Since $A$ is singular but $A+uv^T$ is nonsingular, $u$ lives outside the column space of $A$. The equality $Ax=(1-v^Tx)u$ thus implies that $Ax=0$ and $v^Tx=1$. Therefore $v^T(A+uv^T)^{-1}u=v^Tx=1$.

In terms of matrices, let $A=UV^T$ be a rank factorisation. Since $A$ is singular but $A+uv^T$ is nonsingular, the augmented matrices $\pmatrix{U&u}$ and $\pmatrix{V&v}$ are nonsingular. Therefore \begin{aligned} v^T(A+uv^T)^{-1}u &=v^T\left[\pmatrix{U&u}\pmatrix{V&v}^T\right]^{-1}u\\ &=\left[\pmatrix{V&v}^{-1}v\right]^T\left[\pmatrix{U&u}^{-1}u\right]\\ &=e_n^Te_n=1, \end{aligned} where $e_n=(0,\ldots,0,1)^T$.