= Solution
For a real <Lie group> $G$, its <Lie algebra> is the real <tangent space> $T_eG$ at the identity. Each tangent vector extends uniquely to a left-invariant vector field, and the <commutator> of those fields defines the <Lie bracket>. For a matrix group this bracket is $[X,Y]=XY-YX$.
For a complex <affine algebraic group>, the <Lie algebra of an affine algebraic group> is the complex <Zariski tangent space> $T_eG=\operatorname{Hom}_{\mathbb C}(\mathfrak m_e/\mathfrak m_e^2,\mathbb C)$. Its bracket is the <commutator> of left-invariant derivations. In a faithful matrix realization, one can compute the <tangent space> using <dual numbers>: $X$ belongs to it exactly when $I+\varepsilon X$ satisfies the group equations modulo $\varepsilon^2$. This is a complex <Lie algebra>, whereas <compact groups> such as $U_n$ and $SU_n$ have real <Lie algebras>.
Differentiate $g^*g=I$ at the identity to get $X^*+X=0$. Conversely $e^{tX}$ is unitary for every skew-Hermitian $X$. Differentiating the <determinant> gives $\det(I+tX)=1+t\operatorname{tr}X+O(t^2)$, while $\det e^{tX}=e^{t\operatorname{tr}X}$ proves sufficiency of the trace condition. Finally, $GL_n(\mathbb C)$ is an open subset of all complex matrices, so its <tangent space> has no linear constraint. Therefore
$$
\boxed{\begin{aligned}
\mathfrak u_n&=\{X\in M_n(\mathbb C):X^*=-X\},\\
\mathfrak{gl}_n(\mathbb C)&=M_n(\mathbb C),\\
\mathfrak{su}_n&=\{X:X^*=-X,\ \operatorname{tr}X=0\},\\
\mathfrak{sl}_n(\mathbb C)&=\{X:\operatorname{tr}X=0\}.
\end{aligned}}
$$
Their dimensions are respectively $n^2,n^2,n^2-1,n^2-1$, with real dimensions for the first and third and complex dimensions for the second and fourth. In every case the <Lie bracket> is the matrix <commutator>.
An irreducible complex <affine algebraic group> is connected. It is a <reductive algebraic group> when its <unipotent radical> $R_u(G)$, the largest connected normal unipotent subgroup, is trivial. This is an algebraic condition; in <characteristic zero> it also implies complete reducibility of finite-dimensional <rational representations>, as proved in Question 2.
Examples are <algebraic tori>, $GL_n(\mathbb C)$ and $SL_n(\mathbb C)$, and products of these groups. A diagonal matrix that is unipotent must be the identity, so a torus has no nontrivial unipotent subgroup. For the two matrix groups, let $N$ be a normal unipotent subgroup. The <Kolchin theorem> supplies a nonzero fixed vector in their standard representation. The subspace fixed by $N$ is invariant under the whole group: for $g\in G$, $n(gv)=g(g^{-1}ng)v=gv$. The standard representation is irreducible, so every vector is fixed by $N$; its faithfulness makes $N$ trivial. A normal unipotent subgroup of a product projects to such a subgroup of each factor, proving the product assertion. By contrast the additive group and a nontrivial upper-unitriangular group are not reductive: their <unipotent radicals> are the whole group.
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