1. ## Equivalence

Hello,

Let $E$ a $K$-vector space $(\text{dim}(E){\color{red}\le}\infty)$ and $(u,v)\in L(E)^2$.

Prove that $Id_E-u\circ v$ invertible $\Longleftrightarrow$ $Id_E-v\circ u$ invertible.

2. Originally Posted by Infophile
Hello,

Let $E$ a $K$-vector space $(\text{dim}(E){\color{red}\le}\infty)$ and $(u,v)\in L(E)^2$.

Prove that $Id_E-u\circ v$ invertible $\Longleftrightarrow$ $Id_E-v\circ u$ invertible.

because of the symmetry, we only need to prove $\Longrightarrow.$ let's show $Id_E$ by $1.$

suppose that $1-uv$ is invertible. then there exists $w$ such that $(1 - uv)w = w(1 - uv) = 1,$ which gives us: $uvw = wuv. \ \ \ \ \ (1)$

multiplying (1) by $v$ from the left gives us $vuvw = vwuv.$ therefore $(1-vu)vw = vw(1-uv) = v,$ because $w(1-uv)=1.$

thus $(1-vu)vwu = vu = 1-(1-vu).$ hence: $(1-vu)(1+vwu) = 1.$ this shows that $1+vwu$ is a right inverse of $1-vu.$

this time multilpy (1) by $u$ from the right to get $uvwu=wuvu$ and so $wu(1-vu)=(1-uv)wu=u,$ because $(1-uv)w=1.$

thus: $vwu(1-vu)=vu=1-(1-vu),$ and hence $(1+vwu)(1-vu)=1,$ i.e. $1+vwu$ is also a left inverse of $1-vu.$ Q.E.D.

3. this is true in any ring with unity R, i.e. for any two elements $u,v \in R,$ if the element $1-uv$ is invertible, then $1-vu$ is invertible too and the proof is exactly as i did.

but there's an interesting story behind this. i showed that if $w$ is the inverse of $1-uv,$ then $1+vwu$ would be the inverse of $1-vu.$ professor Tsit Yuen Lam in one

of his books mentions that Kaplansky taught him a way to remember this: since $w$ is the inverse of $1-uv,$ we write: $w=\frac{1}{1-uv}=1+uv+uvuv+uvuvuv + \cdots .$

(geometric series! ) then we'll have: $\frac{1}{1-vu}=1+vu+vuvu + vuvuvu + \cdots = 1 + v(1 + uv + uvuv + uvuvuv + \cdots )u=1+vwu.$

this is anything but a valid solution. it's actually a completely invalid way which gives a correct answer! Kaplansky just wanted to teach his student a little trick!