Prove that $\det(A) \cdot v \, A^{-1} = \det(A+uv) \cdot v \, (A+uv)^{-1}$.

Question

Let $A$ be a $n \times n$ matrix, $u$ a $n \times 1$ matrix and $v$ a $1 \times n$ matrix. If $A$ and $(A+uv)$ are invertible, prove that $$ \det(A) \cdot v \, A^{-1} = \det(A+uv) \cdot v \, (A+uv)^{-1}. $$

I have numerical evidence that this is true, but I cannot prove it. Note that the term on the left does not depend on $u$!

Thank you.

Answer 1

I'll use $v$ here to be $v^T$ in what you have.

The Matrix inversion lemma tells you for $A$ invertible and $u,v$ vectors. Then (provided $u v^T$ doesn't make $A$ singular when added to it), $(A+ u v^T)^{-1} = A^{-1} - \frac{A^{-1} u v^T A^{-1}}{1+ v^T A^{-1} u}$.

The Matrix determinant lemma tells you under the same conditions $det(A+u v^T) =(1+ v^T A^{-1} u) det(A)$.

Plug these lemmas into your conjecture and collect like terms.

Answer 2

I found this proof before falling asleep. By the Matrix inversion lemma, \begin{align*} \dfrac{\det(A+uv)}{\det(A)} v &= \left( 1 + v A^{-1} u \right) v \\ &= v \left( I + A^{-1} u v \right) \\ &= v A^{-1} (A+uv). \\ \end{align*} After multiplying by $(A+uv)^{-1}$ on the right, $$ \dfrac{\det(A+uv)}{\det(A)} v (A+uv)^{-1} = v A^{-1}. $$