Determinant of rank-one perturbations of (invertible) matrices I read something that suggests that if $I$ is the $n$-by-$n$ identity matrix, $v$ is an $n$-dimensional real column vector with $\|v\| = 1$ (standard Euclidean norm), and $t > 0$, then
$$\det(I + t v v^T) = 1 + t$$
Can anyone prove this or provide a reference?
More generally, is there also an (easy) formula for calculating $\det(A + wv^T)$ for $v,w \in \mathbb{K}^{d \times 1}$ and some (invertible) matrix $A \in \Bbb{K}^{d \times d}$?

I solved it. The determinant of $I+tvv^T$ is the product of its eigenvalues. $v$ is an eigenvector with eigenvalue $1+t$. $I+tvv^T$ is real and symmetric, so it has a basis of real mutually orthogonal eigenvectors, one of which is $v$. If $w$ is another one, then $(I+tvv^T)w=w$, so all the other eigenvalues are $1$.
I feel like I should have known this already.  Can anyone provide a reference for this and similar facts?
