There are two statement about a matrix under rank-one updates that I would be grateful if you give me some insightful proofs.
Suppose $A$ be a nonsingular $n \times n$ matrix and $\mathbf{u},\mathbf{v}$ be vectors.
First, Sherman-Morrison Formula states that:
$$(A+\mathbf{u}\mathbf{v}^T)^{-1} = A^{-1} – \frac{A^{-1}\mathbf{u}\mathbf{v}^TA^{-1}}{1+\mathbf{v}^TA^{-1}\mathbf{u}}.$$
(We can prove this by verifying that the RHS multiplied by $A+\mathbf{u}\mathbf{v}^T$ is $I$.)
Second, Matrix Determinant Lemma states that:
$$\det(A+\mathbf{u}\mathbf{v}^T)=\det(A)(1+\mathbf{v}^T A^{-1}\mathbf{u}).$$
(In the proof from wikipedia, we just have to verify some identity again.)
It's easy to verify these proofs but it's not clear to me how to come up with the identity. Are there any other proofs which are not just by multiplication of matrices, and give us some insight ? Or even some informal explanation ?
Best Answer
I should preface this with a disclaimer. It may not be the kind of insight you are looking for and certainly not as insightful as the previous answers, but it is a very direct derivation only requiring the most basic knowledge of linear algebra.
Suppose you wish to solve the following for $\mathbf{x}$:
$$\left(A+\mathbf{uv}^T\right)\mathbf{x}=\mathbf{y}$$ $$A\mathbf{x}=\mathbf{y}-\mathbf{uv}^T\mathbf{x}$$ $$\mathbf{x}=A^{-1}\mathbf{y}-A^{-1}\mathbf{uv}^T\mathbf{x}$$
Notice that $\mathbf{v}^T\mathbf{x}$ is a scalar, let us call it $s$.
So the solution for $\mathbf{x}$ in terms of $s$ is:
$$\mathbf{x}=A^{-1}\mathbf{y}-A^{-1}\mathbf{u}s$$
And solving for $s$:
$$s=\mathbf{v}^T\mathbf{x}=\mathbf{v}^TA^{-1}\mathbf{y}-\mathbf{v}^TA^{-1}\mathbf{u}s$$ $$\left(1+\mathbf{v}^TA^{-1}\mathbf{u}\right)s=\mathbf{v}^TA^{-1}\mathbf{y}$$ $$s=\frac{\mathbf{v}^TA^{-1}\mathbf{y}}{1+\mathbf{v}^TA^{-1}\mathbf{u}}$$
Substituting for $s$ in the solution for $\mathbf{x}$: $$\mathbf{x}=A^{-1}\mathbf{y}-\frac{A^{-1}\mathbf{uv}^TA^{-1}\mathbf{y}}{1+\mathbf{v}^TA^{-1}\mathbf{u}}$$
Or:
$$\mathbf{x}=\left(A^{-1}-\frac{A^{-1}\mathbf{uv}^TA^{-1}}{1+\mathbf{v}^TA^{-1}\mathbf{u}}\right)\mathbf{y}$$
So:
$$\left(A+\mathbf{uv}^T\right)^{-1}=A^{-1}-\frac{A^{-1}\mathbf{uv}^TA^{-1}}{1+\mathbf{v}^TA^{-1}\mathbf{u}}$$