The Matrix determinant lemma states that

$\det(A + uv^T) = det(A)(1 + v^T(A^{-1}u))$

However, I do not understand how do we get the second multiplier here. I was kind of able to understand this proof on wiki right up until the moment where we go from $\det(I + (A^{-1}u)v^T)$ to the multiplier in question. How do we understand that the value of the determinant $\det(I + (A^{-1}u)v^T)$ is $(1 + v^T(A^{-1}u))$? It's not like we know a general formula for the determinant of sum of the identity and something else (or do we? didn't find it)

Thanks in advance.


The wiki have shown the answer, for by $\det (AB)=(\det A)(\det B)$, $\det(I + (A^{-1}u)v^T)$ is also the determinant of $$\begin{bmatrix} I & 0\\ v^T & 1\\ \end{bmatrix} \begin{bmatrix} I+uv^T & u\\ 0 & 1\\ \end{bmatrix} \begin{bmatrix} I & 0\\ -v^T & 1\\ \end{bmatrix}=\begin{bmatrix} I & u\\ 0 & 1+v^Tu\\ \end{bmatrix}$$ where the determinant of RHS is what you want.


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