Show $|A + uv^T| = (1 + v^T A^{-1} u) |A|$

Question

Let $A$ be an invertible square matrix. Prove that

$$ |A + uv^T| = (1 + v^T A^{-1} u) |A|. $$

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My attempt

$|A + uv^T| = \left|A (I + A^{-1}uv^T)\right| = |A| |I + A^{-1} uv^T|$

I'm trying to use the following fact,

$$(I + ab^T)^{-1} = I - \frac{1}{1+b^T a}ab^T$$

but it seems quite tricky because of the $A^{-1}$ part.

Answer 1

You are almost there. For the last step, letting $w:=A^{-1}u$, note $I+wv^T$ has eigenvalues $(1, 1, \dots, 1+v^T w)$. So $$\det(I+A^{-1}uv^T) =\det(I + w v^T) = 1 \times \dots \times (1+v^T w) = (1+v^TA^{-1}u).$$