Let $A_1$ be the matrix algebra consisting of matrices of the form $$ \pmatrix{ \alpha & 0 \\ 0 & \beta\\ }$$ and let $A_2$ be the matrix algebra consisting of matrices of the form $$ \pmatrix{ \alpha & \beta \\ 0 & \alpha\\ }.$$ I want to prove that

these algebras are not isomorphic and that every two-dimensional unital complex Banach algebra is isomorphic to one of them.

I have proved the first part:

If such isomorphism exists, it would have to be multiplicative and bijective. Hence, a idempotent matrix in $A_1$ would be transformed to idempotent matrix in $A_2$. But those algebras have different numbers of idempotent algebras; a contradiction.

Sadly, I have been defeated by the second part. Anyone can help?

  • 1
    $\begingroup$ One observation is that $A_1$ contains no nonzero nilpotent matrices, and $A_2$ does contain nilpotent matrices. Thus, it might be a good idea to consider the two cases whether there exists an element $x\in A$ with $xx=0$ or not. Unfortunately, I don't know how to constuct an isomorphism. $\endgroup$
    – Roland
    Commented Mar 8, 2014 at 12:52

1 Answer 1


Let $B$ be a two-dimensional Banach algebra. Pick $x \in B \setminus \mathbb{C}\cdot e$. Since $\{e,x\}$ is a basis of $B$, we have

$$x^2 = c\cdot e + d\cdot x$$

for some $c,d\in \mathbb{C}$. Now let $\lambda\in\mathbb{C}$ a zero of $z^2 - d\cdot z - c = 0$, then

$$(x-\lambda\cdot e)^2 = x^2 - 2\lambda x + \lambda^2\cdot e = (d-2\lambda)x + (c+\lambda^2)e = (d-2\lambda)(x-\lambda\cdot e) + (c+\lambda d - \lambda^2)e,$$

so setting $y = x-\lambda e$ we have $y^2 = (d-2\lambda)y$ and $y\neq 0$.

If $d - 2\lambda = 0$, i.e. $y^2 = 0$, consider the map

$$(\alpha\cdot e + \beta\cdot y) \mapsto \begin{pmatrix}\alpha &\beta\\ 0 &\alpha\end{pmatrix}.$$

If $d-2\lambda\neq 0$, set $p = \frac{1}{d-2\lambda}y$ to obtain an idempotent $p \notin \{0,1\}$, let $q = e-p$. Then $\{p,q\}$ is a basis of $B$, and you can consider the map

$$(\alpha p + \beta q) \mapsto \begin{pmatrix}\alpha&0\\0&\beta \end{pmatrix}.$$

  • $\begingroup$ I dont entirely understand why to consider those mappings, but answer looks legit. Will think about it later. Thank you. $\endgroup$
    – wroobell
    Commented Mar 8, 2014 at 13:52
  • $\begingroup$ Well, i dont get idea behind this reasoning, i dont know why to consider those mappings. Could you elaborate on that? $\endgroup$
    – wroobell
    Commented Mar 8, 2014 at 20:32
  • $\begingroup$ Those will be the isomorphisms between $B$ and $A_2$ resp. $A_1$. We have a basis, either $\{e,y\}$ with a nilpotent $y$, or $\{p,q\}$ with two independent idempotents. The linearity and bijectivity of the maps is easily verified, then it remains to check that they are multiplicative (and map the unit $e$ to the identity matrix). After that is verified, the proposition follows, every two-dimensional Banach algebra is isomorphic to $A_1$ or $A_2$. $\endgroup$ Commented Mar 8, 2014 at 20:43

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