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I am totally confused with the following proof given in Simmons' Introduction to Topology and Modern Analysis book (page: 312): Screenshot

Suppose $x$ is an arbitrary element in an Banach algebra and $\sigma(x)$ denotes its "spectrum" then Simmon claims:

SS

My Confusion. I am confused at the last line of the proof where he said

$x^n-\lambda 1$ is singular $\iff x-\lambda_i 1$ is singular for at least one $i$

I'm assuming singular means doesn't have a two sided inverse.

'$\implies$' direction is okay: because if every $x-\lambda_i 1$ does have a two sided inverse (i.e., invertible) then their product $x^n-\lambda 1$ also has a two sided inverse.

But how did he claim reverse direction?? I mean if some $x-\lambda_i 1$ is singular (i.e., doesn't have a two sided inverse) it doesn't mean their product also be singular (e.g. here is a counterexample of this fact (?) In this link $a$ and $b$ are not invertible but their product is invertible !)

Where am I making mistake? Can anyone please help me to clarify the above. Thanks.

EDIT Also I am really confused after noticing the link I mentioned. Doesn't that example show that the "reverse implication" in Simmon is not necessarily true?

sigma
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1 Answers1

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Let $A$ be any ring with $1$.

Theorem. Let $a, b \in A$ be elements that commute. Then, $ab$ is invertible iff $a$ and $b$ are invertible.

Proof. $(\Leftarrow)$ is standard.
$(\Rightarrow)$ Let $c$ be such that $(ab)c = 1 = c(ab)$.
The first equation gives $a(bc) = 1$.
The second gives $1 = cab = (cb)a$.
Thus, $a$ has a left and right inverse both. General theory tells us that it is then invertible. Similarly for $b$. $\Box$

Induction gives the following.

Corollary 1. Given pairwise commuting elements $a_1, \ldots, a_n \in A$, the product $a_1 \cdots a_n$ is invertible iff each $a_i$ is.

Taking "singular" to mean "non-invertible", the last result can be rephrased as:

Corollary 2. Given pairwise commuting elements $a_1, \ldots, a_n \in A$, the product $a_1 \cdots a_n$ is singular iff some $a_i$ is.

Now, this applies to your case with $a_i = x - \lambda_i1$, since the elements do commute.

Aryaman Maithani
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  • Here $x$ is an arbitrary element in an Banach algebra. – sigma Jul 22 '21 at 11:47
  • @indrajit: have modified it now – Aryaman Maithani Jul 22 '21 at 12:34
  • Okay I understand actually here “$x-\lambda_i$”’s commutes with each other and so we can consider the subalgebra generated by them and then apply your argument! Right? – sigma Jul 22 '21 at 12:46
  • And the “link” I mentioned in my question is not giving a counter example to your argument because there the elements $a$ and $b$ don’t commute with each other! – sigma Jul 22 '21 at 12:48
  • @indrajit: Please see my modified answer. I have re-written it for two reasons: (1) It makes the process clearer. (2) There was a subtle error in my previous answer. When going from a ring to a subring, I might lose out on inverses. For example, $2 \in \Bbb Z \subset \Bbb Q$ is invertible in $\Bbb Q$ but not in $\Bbb Z$. – Aryaman Maithani Jul 22 '21 at 12:52
  • Also, yes, the counterexample in the link does not apply because here I'm focussing only on elements that commute. – Aryaman Maithani Jul 22 '21 at 12:52
  • Your argument in the previous post was also valid because we were generating polynomial algebra generated by $x$ in the ring which is commutative and contains the identity of the ring. Isn't that? – sigma Jul 22 '21 at 12:57
  • In the general situation we can consider the algebra $\Bbb{C}[a, b]$ which contains the identity also. ($a, b$ commuting elements) – sigma Jul 22 '21 at 12:59
  • @indrajit: I'm not completely sure if my earlier post was valid because: if any of the elements $x^n - 1$ or $x - \lambda_i1$ are invertible in $A$, I cannot see a way of guaranteeing that they are invertible in $\Bbb C[x]$ as well. – Aryaman Maithani Jul 22 '21 at 13:06
  • And I could consider the algebra obtained by adjoining the inverses if they are outside, but then I'd have to argue that it's still commutative. – Aryaman Maithani Jul 22 '21 at 13:07
  • Oh yes you are right! – sigma Jul 22 '21 at 13:17