TITLE: Determinant of rank-one perturbations of (invertible) matrices
QUESTION [18 upvotes]: 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 $A \in \Bbb{K}^{d \times d}$?

REPLY [5 votes]: Here's another proof (cf. Sherman–Morrison formula):
The non-zero eigenvalues of $AB$ and $BA$ are the same. This is straightforward to prove.
Hence the non-zero eigenvalues of $ab^T$ and $b^Ta$ are the same (that is, exactly one non-zero eigenvalue).
Hence the eigenvalues of $I+ ab^T$ are $1+b^Ta, 1,...,1$, and since the determinant is the product of eigenvalues, we have $\det(I+ab^T) = 1+b^Ta$.
In this particular example, $a=tv$, $b=v$, and $\|v\| = 1$, hence $b^Ta = t$, and so $\det(I+t v v^T) = 1+t$.