How to prove $\text{det}(I+xy^{\top}+wz^{\top})=(1+y^{\top}x)(1+z^{\top}w)-(x^{\top}z)(y^{\top}w)$?

Question

Suppose $x,y,z,w$ are vectors in $\mathbb{R}^n$ and $I$ is the identity matrix.

Show that $\text{det}(I+xy^{\top}+wz^{\top})=(1+y^{\top}x)(1+z^{\top}w)-(x^{\top}z)(y^{\top}w)$.

Answer 1

Consider the matrix $$\tag1 \begin{bmatrix} 1+y^Tx & y^Tw \\ z^Tx & 1+z^Tw \end{bmatrix} =I + \begin{bmatrix} y^T\\ z^T \end{bmatrix} \begin{bmatrix} x & w\end{bmatrix} $$ For any $A,B$, we have the equality $\det(I+AB)=\det(I+BA)$. So the determinant in $(1)$ is equal to the determinant of $$ I + \begin{bmatrix} x & w\end{bmatrix}\begin{bmatrix} y^T\\ z^T \end{bmatrix} =I+xy^T+wz^T. $$

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Proof of the equality $\det(I+AB)=\det(I+BA)$.

The matrices $AB$ and $BA$ have the same eigenvalues (padding with zeroes when the sizes are not equal). So $I+AB$ and $I+BA$ have the same eigenvalues, and if the two lists have different length then the remaining eigenvalues are $1$. As the determinant is the product of the eigenvalues, we get $\det(I+AB)=\det(I+BA)$.