Let $a > 0$, $d \in \mathbb{N}$ and define $$ A= I_d + a(v_1 v_1^T + v_2v_2^T) $$ where $I_d$ is the $d$-dimensional identity matrix and $v_1, v_2 \in \{-1, 1\}^d$. What is the determinant of $A$? I wanted to use something like the Matrix Determinant lemma but is is not entirely clear to me how this works when we add two vectors. Any ideas?
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The matrix determinant lemma applies when we add a rank-1 matrix. When we add a matrix of arbitrary rank (that is low relative to the size of the matrix), then the applicable generalization is the Weinstein–Aronszajn identity (AKA the Sylvester identity). In particular, we can write $a v_1v_1^T + av_2v_2^T$ as $aVV^T$, where $V$ is the matrix whose columns are $v_1,v_2$. With that, we have $$ \det(I_d + aVV^T) = \det(I_2 + V^T(aV)) \\= \det \left[I_2 + a\pmatrix{v_1^Tv_1 & v_1^Tv_2\\v_2^Tv_1 & v_2^Tv_2} \right] \\= \det \pmatrix{ad + 1 & v_1^Tv_2 \\ v_1^Tv_2 & ad + 1}\\ = (ad+1)^2 - a^2(v_1^Tv_2)^2\\ = [d^2-(v_1^Tv_2)^2]a^2 + (2d)a + d^2. $$
J.G.
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Ben Grossmann
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1In this context I'd advise writing $v_i\cdot v_j$ instead of $v_i^Tv_j$, but +1 for an excellent approach. – J.G. Jun 23 '20 at 15:34
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@J.G. I disagree – Ben Grossmann Jun 23 '20 at 15:46