The trace form detects semisimplicity over the complex numbers

Left multiplication is $\mathbb{C}$-linear and satisfies $L_{a+\lambda b}=L_a+\lambda L_b$ and $L_{ab}=L_a\circ L_b$, since $cx$ is linear in $c$ with $x$ fixed and $(ab)x=a(bx)$; hence $(\cdot ,\cdot )$ is $\mathbb{C}$-bilinear. Moreover $(ab,c)=\mathrm{tr}(L_{(ab)c})=\mathrm{tr}(L_{a(bc)})=(a,bc)$, and symmetry of the form follows from $(a,b)=\mathrm{tr}(L_a{L}_b)=\mathrm{tr}(L_b{L}_a)=(b,a)$, where the middle equality is the cyclic property of the matrix trace applied to the matrices of $L_a$ and $L_b$ and their composite, with the matrix of a composite given by the product and the trace read in one basis by [F1]. This proves assertion (1).

Assume $A$ is semisimple. If $A=0$ then the empty product gives assertion (3) and the form is nondegenerate vacuously. If $A\mathrm{\ne }0$, Wedderburn-Artin gives a ring isomorphism $A\cong {\prod }_{i=1}^r{\mathrm{M}}_{n_i}(D_i)$ with division rings $D_i$ and $n_i\ge 1$ by [F2]; since $A$ is a $\mathbb{C}$-algebra, the scalar copy of $\mathbb{C}$ lies in the centre of each factor, so each $D_i$ is a finite-dimensional division algebra over $\mathbb{C}$. For $d\in D_i$ the left multiplication $L_d$ on the nonzero finite-dimensional $\mathbb{C}$-space $D_i$ has an eigenvalue $\lambda \in \mathbb{C}$ by [F2], and $L_{d-\lambda }=L_d-\lambda$ is then not injective, so $d-\lambda$, being either $0$ or a unit of the division ring $D_i$, must be $0$; hence $d\in \mathbb{C}$ and $D_i=\mathbb{C}$. Thus $A\cong {\prod }_{i=1}^r{\mathrm{M}}_{d_i}(\mathbb{C})$, and the simple left $A$-modules are the column modules ${\mathbb{C}}^{d_i}$ by [F2]. For $Z\in {\mathrm{M}}_d(\mathbb{C})$ and the basis $\{E_{kl}\}$ of matrix units, $L_Z(E_{kl})=Z{E}_{kl}={\sum }_i{z}_{ik}{E}_{il}$ has no diagonal contribution from the summand indexed by $(k,l)$ other than the $E_{kl}$-coefficient $z_{kk}{E}_{kl}$, so $\mathrm{tr}(L_Z)={\sum }_{k,l}{z}_{kk}=d\mathrm{tr}(Z)$. Hence, under the isomorphism, $(X,Y)={\sum }_i{d}_i\mathrm{tr}(X_i{Y}_i)$; if the first argument is orthogonal to the whole factor $i$, testing against all matrix units $E_{kl}$ of that factor shows $X_i=0$, with $d_i\mathrm{\ne }0$ in $\mathbb{C}$; therefore the form is nondegenerate. This proves the reverse implication of assertion (2) and, with the identification of the simple modules, assertion (3).

Assume conversely that $(\cdot ,\cdot )$ is nondegenerate. Let $J:=J(A)$, which is a two-sided ideal with $A\mathrm{/}J$ semisimple and which is nilpotent by [F3]. For $x\in J$ and $b\in A$ associativity puts $xb\in J$, so $(xb{)}^m=0$ for some $m\ge 1$ and $L_{xb}^m=L_{(xb{)}^m}=0$ by the product rule of step 1.1, that is, $L_{xb}$ is nilpotent; by [F1] it has a strictly upper triangular matrix in some ordered basis, so $(x,b)=\mathrm{tr}(L_{xb})=0$. Since $b\in A$ was arbitrary and the form is nondegenerate, $x=0$. Hence $J=0$ and $A=A\mathrm{/}J$ is semisimple, which is the forward implication of assertion (2).

Assertion (1) is step 1.1, the two directions of assertion (2) are steps 1.2 and 2.1, and assertion (3) is the structure statement proved in step 1.2, including the case $A=0$ as the empty product. All objects constructed are determined by the cited decomposition theorems and by explicit basis computations; no choice principle is invoked, so the item is choice-free.