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I'm trying to understand the alternate proof of the formula $(A+uv^T)^{-1} = A^{-1} - {A^{-1}uv^T A^{-1} \over 1 + v^T A^{-1}u}$ found on wikipedia here.

The closest I can find on stack exchange is the use of the identity but not its proof. Does someone know how $$(I+wv^T)^{-1}=I-\frac{wv^T}{1+v^Tw} \tag{1}$$ is easily proven? I don't see it.

You can mark the question as duplicate if you want but all the proofs are proving the Sherman-Morrison formula, not the intermediate identity given in (1). The two are similar as I see now that someone proved it, but it still is worth showing explicitely how to prove the identity.

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user5389726598465
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2 Answers2

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The idea is to recognize that $v^Tw$ is a scalar. \begin{align}&(I+wv^T) \left(I-\frac{wv^T}{1+v^Tw} \right)\\&= I+wv^T-\frac{wv^T}{1+v^Tw}-\frac{wv^Twv^T}{1+v^Tw} \\&= I+wv^T-\left (\frac{wv^T}{1+v^Tw}+\frac{wv^Twv^T}{1+v^Tw}\right ) \\&= I+wv^T-\frac{w(1+v^Tw)v^T}{1+v^Tw}\\&= I+wv^T-wv^T \\&=I \end{align}

user5389726598465
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Siong Thye Goh
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We can solve the equation $(I+ wv^T) x = y$ directly to obtain $x$:

Suppose $(I+ wv^T) x = x + w v^T x = y$.

Then $v^T x + v^T w v^T x = (1+ v^T w) (v^T x) = v^Ty$, or $v^T x = {v^Ty \over 1 + v^T w}$.

Hence the first equation can be written as $x + {wv^Ty \over 1 + v^T w} = y$ or $x = (I-{wv^T \over 1 + v^T w} ) y$.

copper.hat
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