Why does the recursive decomposition of a finite-dimensional representation into irreducible subspaces terminate?

The block-diagonal matrix form represents the decomposition of the representation $\rho$ into independent subrepresentations on the invariant subspaces $W$ and $W^{\perp}$. When $V = W \oplus W^{\perp}$ and both summands are $G$-invariant, the matrix of $\rho(g)$ in a basis adapted to this direct sum has zero off-diagonal blocks, meaning the action on $W$ does not mix with the action on $W^{\perp}$.

Conditions: $V = W \oplus W^{\perp}$ is a direct sum decomposition.; $W$ and $W^{\perp}$ are both $G$-invariant subspaces.; A basis for $V$ is chosen as the union of bases for $W$ and $W^{\perp}$.

If $W$ is a $G$-invariant subspace and the inner product is $G$-invariant, then for any $y \in W^{\perp}$ and $g \in G$, the vector $\rho(g)y$ remains orthogonal to every $w \in W$. This is shown by rewriting $(\rho(g)y, w)$ as $(y, \rho(g^{-1})w)$.

Conditions: $W \subseteq V$ is a $G$-invariant subspace ($\rho(g)W \subseteq W$ for all $g \in G$).; The inner product $(\cdot, \cdot)$ on $V$ is $G$-invariant ($(\rho(g)v, \rho(g)w) = (v, w)$).; $W^{\perp} = \{y \in V \mid (y, w) = 0 \text{ for all } w \in W\}$.

A group-invariant inner product on the representation space $V$ is defined by the condition that applying the same group element $g$ to both vectors does not change their inner product. Mathematically, this is expressed as $(\rho(g)v, \rho(g)w) = (v, w)$ for all $g \in G$ and all $v, w \in V$.

Conditions: $V$ is a vector space with an inner product $(\cdot, \cdot)$.; $\rho: G \to GL(V)$ is a representation.; The condition must hold for all $g \in G$ and all $v, w \in V$.