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:
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?
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.