1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.