# The invertibility of matrix in Banach algebra

Let $$\mathcal{R}$$ be a Banach algebra with identity element and let $$a = (a_{jk})^n_{j,k=1} \in \mathcal{R}^{n \times n}$$ be a matrix whose entries $$a_{jk} \in \mathcal{R}$$ commute pairwise. Then the fact that $$a$$ is invertible implies that $$\text{det} \ a \in \mathcal{R}$$ is invertible?

The proof is as below:

Now suppose $$a$$ is invertible in $$\mathcal{R}^{n\times n }$$, and $$a^{-1} = c = (c_{jk})^n_{j,k=1} \in \mathcal{R}^{n \times n}$$. It suffices to prove that the elements $$c_{jk}$$ commute pairwise and that they commute with all $$a_{jk}$$ elements, since then the identity $$e = \text{det} (a^{-1} a ) = \det \ a^{-1} \det a$$.

Let $$\mathcal{\zeta}$$ denote the set of all commutative subalgebras of the algebra $$\mathcal{R}$$ containing all entries $$a_{jk}$$ of matrix $$a$$. The set $$\mathcal{\zeta}$$ is partial ordered by inclusion. Thus, by Zorn lemma, $$\mathcal{\zeta}$$ contains at least one maximal element $$\mathcal{U}$$. The commutative subalgebra $$\mathcal{U} \subseteq \mathcal{R}$$ then poessess the following property: if $$a \in \mathcal{R}$$ and if $$ax = xa$$ for all $$x \in \mathcal{U}$$ then $$a \in \mathcal{U}$$. Since $$a_{jk} \in \mathcal{U}$$, for every $$x \in \mathcal{U}$$ the equalities $$a^{-1} x = a^{-1} x a a^{-1} = a^{-1} x = a^{-1} a x a^{-1} = x a^{-1}$$ hold and , hence, $$xc_{jk} = c_{jk} x , (j,k = 1, \cdots, n)$$ for every $$x \in \mathcal{U}$$. This implies that $$c_{jk} \in \mathcal{U}$$.

Note: It is my doubt that, why $$c_{jk}$$ satisfies the equation in bold part? without the invertibility of $$\text{det} \ a \in \mathcal{R}$$, how can we figure out that $$c_{jk}$$ can be represented as the finite sum of composition of $$a_{jk}，j,k = 1,\cdots, n$$ and its inverses, like the form in linear algebra?

Note: This is the step 2 of [MikhlinProssdorf, Singular Integral Operator, P114, Lemma 1.1].

• II don't understand what you mean by "without the invertibility of $\det a$". You've proved that $\det a$ is invertible, so you can use that fact. Jun 4, 2022 at 12:55
• The fact is that $a \in \mathcal{R}^{n \times n }$ is invertible, we aim to prove $\det a \in \mathcal{R}$ is invertible. Jun 4, 2022 at 13:02
• But you have included a proof that $\det a$ is invertible. Is there some step of that proof that you do not understand? Jun 4, 2022 at 13:03
• Yes, the bold part. I can't understand why $c_{jk}$ satisfies the equation. Jun 4, 2022 at 13:07

This is indeed much more subtle than I originally have naively thought. In the current statement of the problem, it's confusing to have $$a\in \mathcal R^{n\times n}$$ and $$a\in\mathcal R$$ in different places. Anyway, here is how I solve it.

Claim: If $$x\in\mathcal R$$ commutes with all entries $$a_{ij}$$ of $$a$$, then it also commutes with all entries $$c_{ij}$$ of $$a^{-1}$$. In particular, all entries of $$a$$ commutes with $$c_{ij}$$'s.

Proof. Let $$X=\text{diag}(x, \cdots, x)\in\mathcal R^{n\times n}$$, note that $$Xa^{-1}a = X = a^{-1}aX = a^{-1} X a$$, we get $$Xa^{-1}a = a^{-1} Xa$$, and mutiply by $$a$$ on the right, there is $$Xa^{-1} = a^{-1}X$$. Now it's straightforward to check $$xc_{ij} = c_{ij}x$$. (We have tried to avoid the scalar multiplication notation for matrices, as the scalar might not commute with the entries.)

In particular, since each $$c_{ij}$$ commutes with all $$a_{ij}$$'s, it must commutes with all of the entries of $$a^{-1}$$, that is $$c_{ij}$$'s commute with each other (this is the part I struggled with). Therefore $$a_{ij}, c_{ij}$$ form a pairwise commutative familly of elements. Now we may use $$\det(ab)=\det(a)\det(b)$$ in a commutative ring to deduce that $$\det(a)$$ is invertible in $$\mathcal R$$.

The subtlety here in my opinion is the following. For example, we could have a matrix $$a\in \mathbb Z^{n\times n}$$ which is not invertible, but became invertible over $$\mathbb Q$$, only because $$\mathbb Q$$ contains $$\det(a)^{-1}$$. This is well-understood in the context of commutative ring extension. However, say we have a unital ring extension $$R\subset S$$ where $$R$$ is commutative and $$S$$ is not. Then $$a\in R^{n\times n}$$ is invertible in $$S^{n\times n}$$ doesn't a priori mean that $$S$$ contains $$\det(a)^{-1}$$, and $$a^{-1} = \det(a)^{-1}adj(a)$$, where $$adj(a)$$ is the adjugate matrix of $$a$$. The above argument shows that it's actually indeed the case.

BTW, this turns out to be purely algebraic as expected, which has little to do with norm and analysis.

• $\mathcal{R}$ may be not commutative. Choose a maximal commutative subalgebra $\mathcal{U}$ for $a$, now, the point is to show why $c$ belongs to the same $M_{n\times n}(\mathcal{U})$. Jun 4, 2022 at 10:57
• I don't get it. Could you elaborate? Jun 4, 2022 at 11:01
• I give the proof in post and emphasize the point. Jun 4, 2022 at 11:21
• Thanks. I started appreciating the problem more and more. Just finished modifying my answer. Jun 4, 2022 at 15:11

The equalities $$xc_{jk}=c_{jk}x$$ follow immediately from the previous sentence in which it is shown that $$a^{-1}x=xa^{-1}$$. The $$(j,k)$$ entry of $$a^{-1}x$$ is $$c_{jk}x$$ and the $$(j,k)$$ entry of $$xa^{-1}$$ is $$xc_{jk}$$.

(Incidentally, the use of Zorn's lemma in the proof is rather ridiculous overkill. You can just directly use this argument with $$x=a_{jk}$$ to show the entries of $$a$$ commute with the entries of $$a^{-1}$$, and then use it again with $$x=c_{jk}$$ to show the entries of $$a^{-1}$$ commute with each other.)

• Now $a$ should belong to $\mathcal{R}^{n \times n }$. Jun 4, 2022 at 13:24
• It can be seen as a multiplication between matrix and scalar, I am confused by the illustration that $a \in \mathcal{R}$ before. Jun 4, 2022 at 13:26
• Here $a$ is considered to represent the $n\times n$ matrix with $a$'s on the diagonal and $0$s off the diagonal. Jun 4, 2022 at 14:15
• $x$ can be like $xI$. Jun 4, 2022 at 14:26