Tracial states and the GNS construction

Question

If a $C^*$-algebra $A$ has a tracial state $\tau$, can we construct a nonzero representation $\pi: A\rightarrow B(H_{\tau})$ such that $\pi(ab)=\pi(ba),\forall a,b \in A$ through the GNS construction?

Answer 1

This is not about GNS. Many C$^*$-algebras lack commutative representations. For instance, if $A=M_n(\mathbb C)$ with $n\geq2$, then any representation is either zero or unitarily equivalent to the identity representation.

Or, if $A=B(H)$, the any representation is either (unitarily equivalent to) the identity, or the quotient map onto the Calkin algebra.

Answer 2

No. Consider $A=M_n(\mathbb C)$ and the standard tracial state $\tau$. Then the GNS representation $\pi:A\to B(H_\tau)$ is injective (an easy exercise), and thus commutators aren't mapped to zero.