 Solution

A simple module is nonzero and has no submodules other than zero and itself. A semisimple module is a direct sum of simple submodules; the zero module is allowed as the empty sum.

For (1)$\Rightarrow$(2), argue by dimension. Choose a minimal nonzero submodule $S$, which is simple, and split $V=S\oplus C$ using (1). The property passes to $C$: for $U\subseteq C$, write $V=U\oplus D$; then $C=U\oplus (C\cap D)$. Apply induction to $C$ and append $S$. The zero-dimensional case starts the induction. The implication (2)$\Rightarrow$(3) is immediate. For (3)$\Rightarrow$(2), repeatedly append a simple submodule not contained in the sum already chosen. Its intersection with that sum is zero by simplicity. Dimension strictly increases, so after finitely many steps their direct sum is $V$.

For (2)$\Rightarrow$(1), write $V={⨁}_i{S}_i$ and fix $U$. Choose a sum $C$ of a maximal subset of these $S_i$ whose sum is disjoint from $U$. If $U+C\ne V$, some $S_j$ is not contained in $U+C$, so simplicity gives $S_j\cap (U+C)=0$. Then $C+S_j$ is a larger direct sum still disjoint from $U$, a contradiction. Hence $V=U\oplus C$.

Condition (1) implies (1*). Conversely suppose (1*) and let $W$ be the sum of all simple submodules. If $W\ne V$, finite dimension supplies a maximal proper submodule $M$ containing $W$. By (1*), $V=M\oplus C$. Projection to the simple quotient $V/M$ identifies the nonzero complement $C$ with a simple module. Thus $C\subseteq W\subseteq M$, contradicting the direct sum. Consequently $W=V$, which is (3). This proves all four conditions are equivalent.

For the concrete complement, $x^2=x$. Orthogonal idempotents $x$ and $1-x$ give $A=Ax\oplus A(1-x)$: the sum follows from $1=x+(1-x)$, and right multiplication by $x$ shows their intersection is zero. In explicit matrix form,The decomposition of $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ uses second-column remainders $b-\lambda a,d-\lambda c$ in $C$.