Why does the existence of a $G$-invariant inner product ensure that the orthogonal complement of a $G$-invariant subspace is also $G$-invariant?

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 $W^{\perp}$.

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 $v, w \in V$.

The decomposition terminates because the dimension of the representation space is finite. At each step, a proper nonzero $G$-invariant subspace $W$ is split off, leaving a complementary invariant subspace $W^{\perp}$.

Conditions: $V$ is a finite-dimensional vector space.; $W$ is a proper nonzero $G$-invariant subspace of $V$.; The inner product is $G$-invariant, ensuring $W^{\perp}$ is also $G$-invariant.