Let $A$ be an (not necessarily unital) algebra over $\mathbb{R}$ or $\mathbb{C}$. If $a\in A$ is an element such that $ab=0$ for all $b\in A$ (or equivalently, $aA=\{0\}$), can we then conclude that $a=0$?

At first sight, it looks like a trivial statement. However, I am not able to prove or counterprove it.

It is clearly true for unital algebras (take $b=1$). I was also able to prove that this statement is true for (complex) C*-algebras (a certain class of algebras):

If $A$ is a C*-algebra, then $A$ admits at least one approximate unit $(u_{\lambda})_{\lambda\in\Lambda}$. By assumption we have $au_{\lambda}=0$ for all $\lambda\in\Lambda$. Taking the limit on both sides yields $a=0$.

Any suggestions would be greatly appreciated. It feels like I'm missing something trivial...

  • $\begingroup$ It is not true for Banach Algebras, I do not have a counter-example in mind, but here is a fact: If it was true, Murphy's book wouldn't state exercise 2.1 as it does $\endgroup$ May 22, 2020 at 21:32

2 Answers 2


We cannot conclude this. Indeed, on any vector space $V$ over any field, one can define multiplication by assigning $vw=0$ for all $v,w\in V$.

Of course, if your algebra $A$ is unital, with unit $e$, then the conclusion does hold. Indeed, $a=ae=0$.

Moreover, there is a more elementary proof that this is true for $C^*$-algebras

Let $A$ be a $C^*$-algebra, and suppose $a\in A$ satisfies $ab=0$ for all $b\in A$. Then $$0=\|aa^*\|=\|a\|^2.$$ Thus $\|a\|=0$, and therefore $a=0$.


Hopefully, I am not saying something stupid.

Pick your favourite algebra $A$.

Let $a \notin A$ be any element, and define $$B= \mathbb{F} a \oplus A$$ where $\mathbb{F}$ is your field.

Now, $B$ becomes an algebra, under the obvious addition and multiplication defined as $$(\alpha a+b)(\beta a +c)=bc$$

And clearly $aB=0$.

  • $\begingroup$ @DavidC.Ullrich Ty, fixed. That was a typo, by construction $aa=0$ too. $\endgroup$
    – N. S.
    May 23, 2020 at 5:40

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