example of Banach algebra which is not a $C^{*}$-algebra for any involution

Consider the Banach algebra with the usual matrix operations and matrix norm, where $I_n$ denotes the identity matrix in $Ma{t}_{n\times n}(ℂ)$.

Claim - $𝒜$ is not a $C^{*}$-algebra (http://planetmath.org/CAlgebra) for any involution $*$.

To prove the above claim we will give a proof of a more general fact about finite dimensional $C^{*}$-algebras, which clearly shows the for a Banach algebra to be a $C^{*}$-algebra for some involution.

Theorem - Every finite dimensional $C^{*}$-algebra is semi-simple, i.e. its Jacobson radical is $\{0\}$.

Proof : Let $ℬ$ be a finite dimensional $C^{*}$-algebra. Let $a$ be an element of $J(ℬ)$, the Jacobson radical of $ℬ$.

$J(ℬ)$ is an ideal of $ℬ$, so $a^{*}a\in J(ℬ)$.

The Jacobson radical of a finite dimensional algebra is nilpotent, therefore there exists $n\in \mathbb{N}$ such that ${(a^{*}a)}^n=0$. Then, by the $C^{*}$ condition and the fact that $a^{*}a$ is selfadjoint,

$$0=\parallel {(a^{*}a)}^{2^n}\parallel ={\parallel a^{*}a\parallel }^{2^n}={\parallel a\parallel }^{2^{n+1}}$$

so $a=0$ and $J(ℬ)$ is trivial. $\mathrm{\square }$

We now prove the above claim.

Proof of the claim: It is easy to see that is the only maximal ideal of $𝒜$. Therefore the Jacobson radical of $𝒜$ is not trivial.

By the theorem we conclude that there is no involution $*$ that makes $𝒜$ into a $C^{*}$-algebra.$\mathrm{\square }$

Remark - It could happen that there were no involutions in $𝒜$ and so the above claim would be uninteresting. That’s not the case here. For example, one can see that $[a_{i,j}]⟶[{\overline{a}}_{2n+1-j,2n+1-i}]$ defines an involution in $𝒜$ (this is just the taken over the other diagonal of the matrix).