# If $A$ is a real or complex algebra and $a\in A$ is such that $ab=0$ for all $b\in A$, then $a=0$?

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...

• 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 May 22, 2020 at 21:32

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 $$aB=0$$.

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