Let $\mathcal{A}$ be a $C^*$-algebra, $I$ a left maximal ideal, then there is a pure state $\rho_I$ s.t. $I = \{x \in \mathcal{A}: \rho_I(x^*x)=0\}$ As the title says, let $\mathcal{A}$ be a $C^*$-algebra and $I$ a left maximal ideal, is it always possible to find a pure state $\rho_I$ such that $I = \{x \in \mathcal{A}: \rho_I(x^*x)=0\}$?
I know this to be true if the $C^*$-algebra $\mathcal{A}$ is unital, in the proof there is no direct mention to its identity element, so is the result valid even for the non-unital case?
Thanks in advance.
 A: Let me outline a series of lemmas that yield what you are asking without using anywhere that $A$ is unital or not. Note that the question only make sense when the maximal ideal is ensured to be closed (for example if the ideal is unimodular, or we refer to a priori closed maximal left ideals). Otherwise, if $I$ is not closed, the question fails trivially, since $\{x:\rho(x^*x)=0\}$ is always closed.
To understand the statements and proofs one needs to be familiar with basic representation theory of $C^*$-algebras, but everything I use here can be found in Murphy's book for example.

Lemma 1: Let $A$ be a $C^*$-algebra and $J\subset A$ a proper, two-sided (closed) ideal. Then there exists an irreducible representation $\phi:A\to B(H)$ such that $J\subset\ker(\phi)$.

Proof of lemma 1: Consider the quotient $C^*$-algebra $A/J$. This is non-trivial since $J$ is proper, so take a pure state $\tau\in PS(A/J)$. Then the associated GNS representation $\pi_\tau:A/J\to B(H_\tau)$ is irreducible. Consider the composition with the quotient map, set $\phi=\pi_\tau\circ q:A\to B(H_\tau)$. Then for $x\in J$ we have $\phi(x)=\pi_\tau(q(x))=\pi_\tau(0)=0$, so $J\subset\ker(\phi)$. But $\phi$ is irreducible by Schur's lemma, since it has a trivial commutant: $\phi(A)'=\pi_\tau(A/J)'=\mathbb{C}1_{H_\tau}$, where we used the fact that irreducibility of $\pi_\tau$ implies triviality of the commutant $\pi_\tau(A/J)'$.

Lemma 2: Let $A$ be a $C^*$-algebra and $I\subset A$ a closed left ideal. Then there exists a pure state $\rho\in PS(A)$ such that for any $x\in I$ we have $\rho(x^*x)=0$.

Proof of lemma 2: Since $I^*$ is a closed right ideal, their intersection $J:= I\cap I^*$ is a closed two-sided ideal. We can thus find an irreducible representation $\phi:A\to B(H)$ such that $J\subset\ker(\phi)$. Let $\xi\in H$ be a cyclic unit vector for the representation and define $\rho:A\to\mathbb{C}$ by $\rho(a):=\langle\phi(a)\xi,\xi\rangle$. It is easily verified that $\rho$ is a state. If $\pi_\rho:A\to B(H_\rho)$ is the associated GNS representation with cyclic unit vector $\xi_\rho$, then we have
$$\langle\pi_\rho(a)\xi_\rho,\xi_\rho\rangle=\rho(a)=\langle\phi(a)\xi,\xi\rangle$$
so $\pi_\rho:A\to B(H_\rho)$ is unitarily equivalent to $\phi:A\to B(H)$. Since the latter is irreducible, $\pi_\rho:A\to B(H_\rho)$ is also irreducible, thus $\rho$ is a pure state. But, if $x\in I$, then $x^*x\in J$, so $\phi(x^*x)=0$, so
$$\rho(x^*x)=\langle\phi(x^*x)\xi,\xi\rangle=0,$$
as we wanted.

Corollary: Let $I\subset A$ be a maximal left ideal. Then there exists a pure state $\rho\in PS(A)$ such that $I=\{x\in A: \rho(x^*x)=0\}$.

Proof of corollary: Employing the above lemma, find a pure state $\rho$ so that $\rho(x^*x)=0$ for all $x\in I$, i.e. $I\subset \{x\in A:\rho(x^*x)=0\}$. But the set on the right hand side is also a left ideal (not equal to $A$ itself since otherwise $\rho=0$, contradicting the fact that $\rho$ is a state), so maximality of $I$ yields equality.
We didn't use anywhere that $A$ is unital or not.
