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Based on this question:$UV$ is invertible if and only if $U$ and $V$ are both invertible

How to modify this wrong statement? Actually, I want to apply that statement to prove the following question:

If $\lambda$ is in spectrum of $A$ then for all $n \in N$, $\lambda^n$ is in spectrum of $A^n$.

Here we need to factor it: $\lambda^n I - A^n = (\lambda I - A)(\lambda^{n-1}I + \lambda^{n-2}A + \dots + \lambda A^{n-2} + A^{n-1}),$ which is easy to check just by distributing and cancelling. (Note that $\lambda I$ always commutes with $A$.)

If $\lambda^n I - A^n$ has an inverse, then so does $\lambda I - A$.

Update: our statement: Let $U, V$ be linear operator on normed space $X$. Assume that $UV=VU$. Then $UV$ is invertible if and only if $U$ and $V$ are both invertible.

amWhy
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Hermi
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  • Note that in the question you linked in the linked question, we also have $UV=VU$. – Brian Moehring Dec 02 '23 at 23:25
  • For the question in the post, I think you forgot the note $\lambda I$ commutes with $A$. This implies that $ (\lambda^{n-1}I + \lambda^{n-2}A + \dots + \lambda A^{n-2} + A^{n-1})$ and $(\lambda I - A)$ commutes, and you can thus construct the inverse of $(\lambda I - A)$ assumed the inverse of $ (\lambda^{n-1}I + \lambda^{n-2}A + \dots + \lambda A^{n-2} + A^{n-1})$. – Tri Dec 03 '23 at 00:02
  • @Tri But what is the inverse of $(\lambda^{n-1}I+\dots A^{n-1})$? That is a problem... So we can only get the left inverse and right inver of $(\lambda I-A)$ as comments. – Hermi Dec 03 '23 at 00:33
  • @Hermi sorry, I copied the wrong one. I meant the inverse of $\lambda^n I -A^n$ - you can call it $B$ and then $X=(\lambda^{n-1}I + \lambda^{n-2}A + \dots + \lambda A^{n-2} + A^{n-1})B$ is the inverse of $\lambda I-A$. We don't need to find the inverse of $(\lambda^{n-1}I + \lambda^{n-2}A + \dots + \lambda A^{n-2} + A^{n-1})$ – Tri Dec 03 '23 at 00:59

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The correct lemma is this:

Let $X$ be a vector space and let $U,V$ be endomorphisms on $X$ such that $UV = VU$. Then $UV$ is invertible if and only if $U$ and $V$ are both invertible.

(This can be generalized a little more, but this suffices for the current situation you find yourself in)

Brian Moehring
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  • And instead of $\mathrm{End}(X)$ we can work in any ring. – Martin Brandenburg Dec 02 '23 at 23:39
  • @MartinBrandenburg Right. In fact, I suppose we can work in any monoid. – Brian Moehring Dec 02 '23 at 23:45
  • What do you mean $U, V$ be endomorphisms? Can you please write the statement in sense of functional analysis? Let $X$ be a normed space or Banach space and $U, V$ be linear operator? Thank you! Also, if it is possible, can you please write a proof of that? – Hermi Dec 03 '23 at 00:34
  • So the statement is that let $U, V$ be linear operator on normed space $X$. Assume that $UV=VU$. Then $UV$ is invertible if and only if $U$ and $V$ are both invertible? – Hermi Dec 03 '23 at 00:36
  • For one side proof: assume that $U$ and $V$ are invertible, then the inverse of $UV$ is $S:=V^{-1}U^{-1}$. Indeed, $UVS=I=S(UV)$. But assume that $UV$ is invertible, how to show $U$ and $V$ are both invertible? – Hermi Dec 03 '23 at 00:37
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    @Hermi Let $S$ be the inverse of $UV$. Then $U(VS) = (UV)S = I = S(UV) = S(VU) = (SV)U$. As $U$ has both a left inverse and a right inverse, it is invertible. – Brian Moehring Dec 03 '23 at 00:42
  • @BrianMoehring But for the definition of inverse operator, we say $U$ is invertible if there exists operator $S$ so that $SU=US=I$. But here our left inverse is not same as the right inverse. How to deal with that? – Hermi Dec 03 '23 at 03:25
  • @Hermi You should study some algebra. A lot of these issues are standard exercises in a linear algebra or group theory class. In particular, if $S_LU=US_R = I$ is a two-sided identity element, then $$S_L = S_LI=S_L(US_R)=(S_LU)S_R = IS_R=S_R.$$ As I said in the previous comment, "as $U$ has both a left inverse and a right inverse, it is invertible" – Brian Moehring Dec 03 '23 at 04:00
  • @BrianMoehring Thank you! – Hermi Dec 03 '23 at 04:02