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Math Help - Equivalence

  1. #1
    Junior Member Infophile's Avatar
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    Smile 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.

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  2. #2
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    Quote Originally Posted by Infophile View Post
    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.
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  3. #3
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    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!
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