How can I verify $(I-UV^T)^{-1}=I+(UV^T)/(1+V^T U)$?

Question

Let $U = [u_1, u_2,\dots ,u_n]^T$ and $V= [v_1,v_2\dots,v_n]^T$ two column vectors. Let $A$ be the matrix $UV^T$.

How can I verify $(I-A)^{-1}=I+(UV^T)/(1+V^T U)$?

I couldn't approach any idea...

Answer 1

We will show that if $s:=V^TU=\sum_{k=1}^n v_k u_k\not=1$ and $A:=UV^T$ then $I-A$ is invertible and $$(I-A)^{-1}=I+(UV^T)/(1-V^T U)$$ (there should be a minus sign at the denominator in the right-hand side).

We first note that $A=[u_iv_j]$. Hence $$A^2=\left[\sum_{k=1}^n (u_iv_k)(u_kv_j)\right]=\left[u_iv_j\sum_{k=1}^n v_ku_k\right]=s[u_i v_j]=sA.$$ Let us assume that $s\not=1$ $$B:=\left(I+\left(UV^T\right)/\left(1-V^TU\right)\right)=I+\frac{A}{1-s}$$ then $$(I-A)B=(I-A)\left(I+\frac{A}{1-s}\right)=I-A+\frac{A}{1-s}-\frac{A^2}{1-s}\\ =I-A+\frac{A}{1-s}-\frac{sA}{1-s}=I+\frac{-(1-s)A+A-sA}{1-s}=I.$$ In a similar way, we verify that $B(I-A)=I$. Hence $B=(I-A)^{-1}$.

Answer 2

To verify it, show that

$$\left(I-A\right)\left(I+\left(UV^T\right)/\left(1+V^TU\right)\right)=I$$