TITLE: Proving the relation $\det(I + xy^T ) = 1 + x^Ty$
QUESTION [18 upvotes]: Let $x$ and $y$ denote two length-$n$ column vectors. Prove that $$\det(I + xy^T ) = 1 + x^Ty$$

Is Sylvester's determinant theorem an extension of the problem? Is the approach the same?

REPLY [5 votes]: An argument without rigorous proof is:
(1) Note that $xy^T$ is a rank 1 matrix with eigenvalues zeros and $x^Ty$.
(2) The eigenvalues of $I+xy^T$ are ones and $1+x^Ty$. 
(3) Since the product of eigenvalues produces the determinant, then
$$det(I+xy^T)=1+x^Ty.$$