If element in $C^{\ast}$ algebra acts as multiplication by scalar is it equal to this scalar? Suppose $A$ is a non-unital $C^{\ast}$-algebra. Does there exist a $a\in A$, with the property that $ab=\lambda b$ for each $b\in B$, with $\lambda$ some complex scalar?
 A: Denote $\widetilde{A}=A\oplus\mathbb{C}$ the unitization of $A$. Then the following works, endow the bounded linear operators $B(\widetilde{A})$ with the operator norm, $||\cdot||_{op}$. Note that
$$||a-\lambda||_{op}=\sup\limits_{b\in\widetilde{A}\,\,\,||b||=1}||ab-\lambda b||=\sup\limits_{b\in\widetilde{A}\,\,\,||b||=1}||\lambda b-\lambda b||=0.$$
As the operator norm is non-degenerate we conclude that $a=\lambda$, which would be absurd as $A$ is non-unital.
A: This is impossible, unless $a = 0$.
We claim that there must be a unique scalar $\lambda$ such that $ab = \lambda b$ for all $b \in A$. Indeed, clearly when $b,c \in B$ are linearly dependent, if $ab = \lambda b$ then $ac = \lambda c$. Now suppose that $b_1,b_2$ are linearly independent (which if we assume that the algebra is non-unital, such elements exist since the algebra is necessarily infinite-dimensional) and $ab_1 = \lambda_1 b_1, ab_2 = \lambda_2 b_2$, and $a(b_1 + b_2) = \lambda(b_1 + b_2)$. But since multiplication on the left by a fixed element is linear,
$$ \lambda_1 b_1 + \lambda_2 b_2 = \lambda b_1 + \lambda b_2. $$
Rearranging we get
$$ (\lambda_1 - \lambda)b_1 + (\lambda_2 - \lambda)b_2 = 0, $$
which implies that $\lambda_1 = \lambda_2 = \lambda$ by the linear independence of $b_1,b_2$.
Okay, but now there is a fixed $\lambda$ such that $ab = \lambda b$ for all $ b\in A$. If $\lambda \neq 0$, then $\frac{1}{\lambda}ab = b$ for all $b \in A$, giving that $A$ contains a left identity, call it $u = \frac{1}{\lambda}a$. Notice that $u^*$ is a right-identity for $A$ since $bu^* = (ub^*)^* = (b^*)^* = b$. Moreover since $u$ is a left-identity and $u^*$ is a right-identity,
$$u = uu^* = u^*. $$
So $A$ is unital with unit $u$, which is a contradiction. So $\lambda = 0$.
