How to calculate $(I - xy^T)^{-1}$?

Question

Let $x, y \in \mathbb R^n$ and suppose that $x^T y \neq 1$, compute $(I - xy^T)^{-1}$.

Really don't know how to work it out. Thanks!

Answer 1

When I see something like $(1-z)^{-1}$ I think of $1+z+z^2+\ldots\ \ $ Let's try that here without worrying about niceties like convergence of the sum. Note that $(xy^T)^n=x(y^Tx)^{n-1}y^T$ so we try $$(1-xy^T)^{-1}=\sum_{i=0}^\infty (xy^T)^i=1+x\left(\sum_{i=0}^\infty(y^Tx)^i\right)y^T=1+\frac 1{1-y^Tx}xy^T$$ If you multiply this by $1-xy^T$ you get $1$ as desired.

Answer 2

According to Cayley Hamilton, $(I-xy^T)^{-1}$ is a polynomial in $xy^T$, that is $a_0I+a_1xy^T+\cdots+a_{n-1}(xy^T)^{n-1}$. Since $(xy^T)^2=(y^Tx)(xy^T)$, we deduce that $(I-xy^T)^{-1}$ is in the form $aI+bxy^T$.

From $(aI+bxy^T)(I-xy^T)=I$, we deduce, by identification, that $a=1,b=1/(1-y^Tx)$.