Existence of non- normal element element in a non commutative C$^*$-algebra

Question

I'm not familiar with lots of examples of non commutative C$^*$-algebras, but there are $M_n(\mathbb{C})$, and $B(H)$. These have a non trivial non-normal element.

My question is : what about the general case ? Does there always exists a non-normal element in a non-commutative C$^*$-algebra?

Answer 1

Suppose $A$ is a non-commutative $C^*$-algebra. Since $A$ is the span of its self-adjoint elements, there exist $x,y\in A$ such that $x$ and $y$ are self-adjoint and $[x,y]=xy-yx\neq0$. Put $z=x+iy$. Then $z^*=x-iy$, and $$[z,z^*]=-2i[x,y]\neq0.$$ Thus, $z$ is non-normal.

Answer 2

It is easy to see that if all elements in $A$ are normal, then $A$ is commutative.

Assume that all elements in $A$ are normal. For any $a,b\in A$ you would have $$(a+b)(a+b)^*=(a+b)^*(a+b).$$ Using that $a,b$ are normal, this simplifies to $$\tag1\operatorname{Re} a^*b=\operatorname{Re}ab^*.$$ If we use $(1)$ for $ia$ and $b$, with $ab$ selfadjoint, we get $$\tag2 i\operatorname{Im}(ab)=\operatorname{Re}(-ia^*b)=\operatorname{Re}((ia)^*b)=\operatorname{Re}(iab^*)=-i\operatorname{Im}(ab). $$ So $\operatorname{Im}(ab)=0$, implying that $ab$ is selfadjoint. Then $ba=(ab)^*=ab$, and all selfadjoint elements commute. As these span the algebra, the whole algebra is commutative.