Evaluate binary determinant

Question

Let $\alpha=(a_1, \dots, a_n), \beta=(b_1, \dots, b_n) \in \mathbb{F}_2^n$. Let $E$ denote the identity matrix. I am trying to evaluate determinant $\det(E + \alpha^T\beta) $. Some experiments show that it will be equal to $1+\alpha\beta^T$.

Can anybody help with proof or may be give some hints how to apply an induction principle?

Answer 1

$$ \det \left( {\bf I} + {\bf a} {\bf b}^\top \right) = \det \begin{bmatrix} {\bf I} & - {\bf a} \\ {\bf b}^\top & 1 \end{bmatrix} = 1 + {\bf b}^\top {\bf a} $$

depending on which Schur complement one uses. Some call it the Weinstein-Aronszajn identity.

Answer 2

Let's assume $\alpha$ nor $\beta$ are zero, since then the result is clear. I prefer them to be column vectors instead of row vectors.

Assuming $\beta^T\alpha \neq 0$ we can show that $I_n + \alpha \beta^T$ is diagonalizable with eigenvalues 1 (with geometric multiplicity $n-1$) and $1+\alpha^T\beta$ (with geometric multiplicity 1). The result then follows because the determinant is the product of the eigenvalues.

The space of vectors perpendicular to $\beta$ has dimension $n-1$, and any such nonzero vector $\mathbf{x}$ is a 1-eigenvector for $I_n+\alpha \beta^T$ because $(I_n+\alpha\beta^T)\mathbf{x} = \mathbf{x}+\alpha\beta^T\mathbf{x} = \mathbf{x}$. Thus the multiplicity of 1 as an eigenvector is at least $n-1$.

The vector $\alpha$ is an eigenvector with eigenvalue $1+\alpha^T\beta$ as $(I_n+\alpha\beta^T)\alpha = \alpha + \alpha (\beta^T\alpha) = (1+\beta^T\alpha) \alpha = (1+\alpha^T\beta)\alpha$. Since the sum of the dimensions of the eigenspaces is at most $n$, it must be that the 1-eigenspace had dimension $n-1$ and the $(1+\alpha^T\beta)$-eigenspace had dimension 1.

Finally, consider the case $\alpha^T\beta = 0$. Let $v_1, \ldots, v_n$ be a basis for the space of column vectors in which $v_2, \ldots, v_n$ are a basis for the subspace of vectors orthogonal to $\beta$ and hence 1-eigenvectors for $1+\alpha\beta^T$. We have $(1+\alpha\beta^T)v_1 = v_1 + (\beta^Tv_1) \alpha$ which is $v_1$ plus a vector orthogonal to $\beta$ so the matrix of the linear map given by multiplication by $I_n+\alpha\beta^T$ with respect to this basis is lower triangular with 1s on the diagonal and therfore has determinant 1.