# Element $a$ of a C*-algebra such that $f(a^*a) = 0$ but $f(aa^*) \neq 0$ for some positive linear functional $f$

Could somebody please give me an example, assuming one exists, of an element $a$ of a C*-algebra $A$ such that $f(a^*a) = 0$ but $f(aa^*) \neq 0$ for some positive linear functional $f:A \to \mathbb{C}$? I tried to prove there was no example, but that didn't seem to work out so now I'm guessing it's false. Thanks in advance.

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Let $e_0=(1,0,0,0,0,\ldots)\in\ell^2$, let $f:B(\ell^2)\to\mathbb C$ be defined by $f(T)=\langle Te_0,e_0\rangle$, and let $S\in B(\ell^2)$ be the "backward shift" defined by $S(a_0,a_1,a_2,a_3,\ldots)=(a_1,a_2,a_3,a_4,\ldots)$. Then $f(SS^*)=1$ but $f(S^*S)=0$.
Similar but simpler: $A=M_2(\mathbb C)$, $f(T)=\left\langle T\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}1\\0\end{bmatrix} \right\rangle$, $a=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.
Aghhh I was trying to make that exact example $A=M_2(\mathbb{C})$, $a=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ work, but the only positive linear functional I could think of was the trace, which didn't work. – Mike F Dec 5 '11 at 5:36
So, generally, if $A = B(H)$ for some Hilbert space, then $T \mapsto \langle Tx,x \rangle$ is always an example of a positive linear functional when $x \in H$? I didn't realize they were so easy to conjure up. By the way, are those all of the positive linear functionals when $A = B(H)$? – Mike F Dec 5 '11 at 5:39
@Mike: $T\mapsto\langle Tx,x\rangle$ is sometimes called a positive vector functional, and a vector state if $\|x\|=1$. No, not all positive linear functional on $B(H)$ have this form. Perhaps the simplest way to see this is to note that a sum of two positive linear functionals is a positive linear functional, but not typically a vector functional itself. The trace on $M_2(\mathbb C)$ is a sum of two vector functionals, but is not a vector functional. The positive linear functionals on $B(H)$ that are finite sums of positive vector functionals are precisely those that are continuous... – Jonas Meyer Dec 5 '11 at 5:57
...with respect to the weak operator topology on $B(H)$. Allowing infinite sums gives the functionals continuous with respect to the ultraweak operator topology (which coincides with the weak* topology), also known as positive normal functionals. There are also nonnormal functionals when $H$ is infinite dimensional. E.g., there are nonzero positive linear functionals on $B(H)$ that vanish on the ultraweakly dense subalgebra of compact operators (as can be obtained using Hahn-Banach). – Jonas Meyer Dec 5 '11 at 6:00
Righhht the trace on $M_2(\mathbb{C})$ is not of of that form. I had momentarily bamboozled myself into thinking $x=(1,1)$ worked. Also I had some kind of vague idea along the lines of 1) take a positive linear functional $f$ on $A$, 2) carry out the GNS construction to get a representation $\pi$ of $A$ on an $H$ that is a quotient of $A$ 3) OK now $f(a) = \langle \pi (a) x,x \rangle$ where $x$ is the class of the unit (assuming such exists) in $H$, so $f$ is "sort of" a coming from a vector, but the Hilbert space might change... – Mike F Dec 5 '11 at 6:19