Efficient computation of determinant of identity minus outer product

Question

Suppose $\hat{x}\in\mathbb{R}^d$ is a normalized vector and $\alpha\in(0, 1)$ a scalar. Is there a way to compute the determinant of the following matrix quickly? $$ I_d - \alpha\hat{x}\hat{x}^\top $$ To me this structure seems to have a lot of structure so I'd expect some trick to do the job.

Answer 1

As discussed in the comments, $\hat{x}\hat{x}^\top$ is a rank $1$ matrix, so all but one of the eigenvalues is $0$. The remaining eigenvalue can be observed thusly: $$(\hat{x}\hat{x}^\top)\hat{x} = (\hat{x}^\top\hat{x})\hat{x} = \|\hat{x}\|^2\hat{x} = \hat{x},$$ since $\hat{x}$ is normalised. Thus, the eigenvalues are $0$ with multiplicity $n - 1$ (where $\hat{x} \in \Bbb{R}^n$) and $1$ with multiplicity $1$.

Note that $\alpha \hat{x}\hat{x}^\top$ and $I - \alpha \hat{x}\hat{x}^\top$ have the same eigenvectors, with respective eigenvalues $\alpha, 0, 0, \ldots, 0$ and $1 - \alpha, 1, 1, \ldots, 1$. The determinant of the latter is therefore $$(1 - \alpha) \cdot 1^{n - 1} = 1 - \alpha.$$

EDIT: An even quicker proof involves this identity. Let $A = i\sqrt{\alpha}\hat{x}$ and $B = i\sqrt{\alpha}\hat{x}^\top$. Then $$\det(I_n - \alpha\hat{x}\hat{x}^\top) = \det(I_n + AB) = \det(I_1 + BA) = \det(I_1 - \alpha\hat{x}^\top \hat{x}).$$ The latter is the determinant of the $1 \times 1$ matrix $(1 - \alpha)$, which is $1 - \alpha$, thus giving us the result again.