# Inverse of matrix sum of identity and outer product

So before we begin, I already know the answer. I'm just having difficulty figuring out the steps for finding it.

Given $u,v \in \mathbb{R}^{n}$, I want to show that $$(I+uv^{T})^{-1}= I - \frac{uv^{T}}{1+v^{T}u}$$

I know from Inverse of the sum of matrices that this is the answer, and since both O(u)=O(v)=n that $Tr(uv^{T}) = v^{T}u$. It's just a matter of getting from point A to point B.

• Your expression for the inverse isn't clear...in fact, not much is. You should learn quickly how to properly write mathematics in this site. – Timbuc Sep 29 '14 at 15:28
• Matrix $B$ is the inverse of matrix $A$ if and only if... – leo Sep 29 '14 at 16:00
• If $u,v\in\Bbb R^n$ then $uv^T$ is a number, or do you mean $u,v\in\Bbb R^{n\times 1}$. – leo Oct 7 '14 at 3:10

A matrix $A$ of rank${}\leq1$ like $uv^T$ has eigenvalue $0$ of geometric multiplicity at least $n-1$, and the remaining eigenvalue is the trace of$~A$, here $v^Tu$ since $\def\tr{\operatorname{tr}}\tr(uv^T)=\tr(v^Tu)$. It follows that the polynomial $X(X-\tr A$) annihilates $A$, that is $A(A-\tr(A)I)=0$.
Writing $t=\tr(A)$ for simplicity, the relation $X(X-t)=0$ implies $(X+1)(t+1-X)=t+1$, so provided that $t\neq-1$, the image of $X+1$ is invertible in the quotient ring $K[X]/(X(X-t))$, with inverse $\frac{t+1-X}{t+1}=1-\frac X{t+1}$. Now substituting back $A=uv^T$ for $X$ and $v^Tu$ for $t$ gives $$(I+uv^T)^{-1}=I-\frac1{1+v^Tu}uv^T$$
• Yes, thanks. Though in the context of arithmetic with square matrices it is fairly natural to interpret $I$ as just a manifestation of (the multiplicative ring unit) $1$. – Marc van Leeuwen Mar 31 '17 at 20:02