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$\DeclareMathOperator\Spec{Spec}$Let $A$ be a finite dimensional $*$-algebra over $\mathbb C$.
(Namely, an associate algebra equipped with an involution $*:A\to A$ satisfying $(ab)^*=b^*a^*$ and $(\lambda a)^*=\bar\lambda a^*$.)

Assume that for $\forall a\in A$ we have $\Spec(a^*a)\subset\mathbb R_+$.
Does it follow that $A$ is a C*-algebra?

Here, the spectrum $\Spec(x)$ of an element $x$ is the set of scalars $\lambda\in \mathbb C$ such that $x-\lambda$ is not invertible.

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    $\begingroup$ I suppose you mean "does there exist a norm such that $A$ is a C$^*$-algebra?". $\endgroup$
    – Chris Ramsey
    Commented Aug 17, 2020 at 20:33
  • $\begingroup$ @Chris: the norm is completely determined by the underlying *-algebra (if it exists); the grammar of the question is fine. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 17, 2020 at 20:45
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    $\begingroup$ @QiaochuYuan Yes, I agree with you. I was making the point that the algebraic structure can be paired with many other norms which do not yield C$^*$-algebras. $\endgroup$
    – Chris Ramsey
    Commented Aug 17, 2020 at 20:52
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    $\begingroup$ @Ruy Indeed, the algebra $\mathbb C[x]/x^2$ with conjugation $(a+bx)^*:=\bar a+\bar b x$ is a counterexample. You may post it as an answer and I will accept it. $\endgroup$
    – André Henriques
    Commented Aug 17, 2020 at 21:01

1 Answer 1

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Let $V$ be a complex vector space equipped with an involutive anti-linear star operation (e.g. a C*-algebra whose multiplication has been forgotten). Equip $V$ with the identically zero multiplication, namely $xy=0$ for all $x$ and $y$ in $V$. Then the unitization of $V$ is a counter-example. In fact, every element $a$ of $V$ is nilpotent so $\text{spec}(a) = \{0\}$. Consequently the spectrum of any element of the form $a-\lambda$ is $\lambda$ from where one easily checks the required condition.

However $a^*a=0$ for every $a$ in $V$, so $\tilde V$ cannot possibly be a C*-algebra.

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