[proofplan]
Split into two cases. If $V \not\cong W$, Schur's lemma forces every $G$-homomorphism $V \to W$ to be zero, so $\operatorname{Hom}_G(V, W) = 0$. If $V \cong W$, fix a $G$-isomorphism $\theta_0: V \to W$ and show that every other non-zero $\theta \in \operatorname{Hom}_G(V, W)$ is a scalar multiple of $\theta_0$ — composing $\theta_0^{-1} \theta$ gives a $G$-endomorphism of $V$, which by part 2 of Schur's lemma is a scalar multiple of $\iota_V$, hence $\theta = \lambda \theta_0$ for some scalar $\lambda$. This shows $\operatorname{Hom}_G(V, W) = \mathbb{C}\theta_0$, a one-dimensional $\mathbb{C}$-vector space.
[/proofplan]
[step:Show $\operatorname{Hom}_G(V, W) = 0$ when $V \not\cong W$]
Suppose $V \not\cong W$ as $G$-spaces. Let $\theta \in \operatorname{Hom}_G(V, W)$. By part 1 of [Schur's Lemma](/theorems/2414), applied to the irreducible $G$-spaces $V$ and $W$ and the $G$-homomorphism $\theta$, either $\theta = 0$ or $\theta$ is a $G$-isomorphism. The latter would give $V \cong W$, contradicting our hypothesis. Hence $\theta = 0$.
So every element of $\operatorname{Hom}_G(V, W)$ is zero, i.e., $\operatorname{Hom}_G(V, W) = 0$, and $\dim_\mathbb{C} \operatorname{Hom}_G(V, W) = 0$.
[/step]
[step:Fix a $G$-isomorphism $\theta_0: V \to W$ in the case $V \cong W$]
Now suppose $V \cong W$ as $G$-spaces. By definition there exists a $G$-isomorphism
\begin{align*}
\theta_0: V \to W,
\end{align*}
i.e., a bijective $\mathbb{C}$-linear map with $\theta_0(gv) = g\theta_0(v)$ for all $g \in G$, $v \in V$. Fix such a $\theta_0$ for the remainder of the proof.
In particular, $\theta_0 \in \operatorname{Hom}_G(V, W)$ and $\theta_0 \neq 0$ (it is bijective and $V \neq 0$). Hence $\dim_\mathbb{C} \operatorname{Hom}_G(V, W) \geq 1$.
[/step]
[step:Show every $\theta \in \operatorname{Hom}_G(V, W)$ is a scalar multiple of $\theta_0$]
Let $\theta \in \operatorname{Hom}_G(V, W)$ be arbitrary.
Define
\begin{align*}
\phi: V &\to V \\
v &\mapsto \theta_0^{-1}\bigl(\theta(v)\bigr).
\end{align*}
This is a composition of two $\mathbb{C}$-linear maps, hence $\mathbb{C}$-linear. It is a $G$-homomorphism: for $g \in G$ and $v \in V$,
\begin{align*}
\phi(gv) = \theta_0^{-1}(\theta(gv)) = \theta_0^{-1}(g\theta(v)) = g\theta_0^{-1}(\theta(v)) = g\phi(v),
\end{align*}
using $G$-equivariance of $\theta$ and of $\theta_0^{-1}$. ($G$-equivariance of $\theta_0^{-1}$ follows from that of $\theta_0$: if $\theta_0(v) = w$, then $\theta_0(gv) = gw$, so $\theta_0^{-1}(gw) = gv = g\theta_0^{-1}(w)$.)
So $\phi \in \operatorname{End}_G(V)$. Since $V$ is irreducible and the base field is $\mathbb{C}$, which is algebraically closed, part 2 of [Schur's Lemma](/theorems/2414) applies and gives $\phi = \lambda \iota_V$ for some $\lambda \in \mathbb{C}$.
Substituting back,
\begin{align*}
\theta_0^{-1} \circ \theta = \lambda \iota_V \quad \implies \quad \theta = \theta_0 \circ (\lambda \iota_V) = \lambda \theta_0,
\end{align*}
the second equality following from $\mathbb{C}$-linearity of $\theta_0$ (composition with $\lambda \iota_V$ scales by $\lambda$).
So every $\theta \in \operatorname{Hom}_G(V, W)$ is of the form $\lambda \theta_0$ for some $\lambda \in \mathbb{C}$. Conversely, every $\lambda \theta_0$ is a $\mathbb{C}$-linear combination of $G$-homomorphisms (a single one), hence a $G$-homomorphism. Therefore
\begin{align*}
\operatorname{Hom}_G(V, W) = \{\lambda \theta_0 : \lambda \in \mathbb{C}\} = \mathbb{C}\theta_0,
\end{align*}
a one-dimensional $\mathbb{C}$-vector subspace of $\operatorname{Hom}_\mathbb{C}(V, W)$. Hence $\dim_\mathbb{C} \operatorname{Hom}_G(V, W) = 1$.
[guided]
The proof has two cases — non-isomorphic and isomorphic — handled by the two parts of Schur's lemma.
**Case $V \not\cong W$.** Schur part 1 says every $G$-homomorphism between irreducibles is either zero or an isomorphism. The "isomorphism" branch is forbidden by hypothesis (it would give $V \cong W$), so every $G$-homomorphism is zero. The Hom space is $\{0\}$, dimension $0$.
**Case $V \cong W$.** Fix one $G$-isomorphism $\theta_0$. The strategy is: every $G$-homomorphism $\theta: V \to W$, when "composed back" via $\theta_0^{-1}$, becomes a $G$-endomorphism of $V$. Schur part 2 (using algebraic closedness of $\mathbb{C}$) says every $G$-endomorphism of an irreducible is a scalar multiple of the identity. Hence $\theta_0^{-1}\theta = \lambda \iota_V$, so $\theta = \lambda \theta_0$.
**Why is $\theta_0^{-1}$ a $G$-homomorphism?** A $G$-isomorphism is a bijective $G$-homomorphism; its inverse is automatically a $G$-homomorphism. To see this, suppose $\theta_0(gv) = g\theta_0(v)$. Set $w := \theta_0(v)$, so $v = \theta_0^{-1}(w)$. Apply $\theta_0^{-1}$ to both sides of $\theta_0(gv) = g\theta_0(v)$:
\begin{align*}
gv = \theta_0^{-1}(g\theta_0(v)) = \theta_0^{-1}(gw).
\end{align*}
Substituting $v = \theta_0^{-1}(w)$,
\begin{align*}
g\theta_0^{-1}(w) = \theta_0^{-1}(gw),
\end{align*}
which is $G$-equivariance of $\theta_0^{-1}$.
**Why is $\mathbb{C}$ — i.e., algebraic closedness — needed?** The hypothesis $\mathbb{F} = \mathbb{C}$ is consumed in the application of Schur part 2, which requires algebraic closedness to extract an eigenvalue. Over a non-algebraically-closed field like $\mathbb{R}$, the dimension formula fails: $\operatorname{Hom}_G(V, V)$ may be a division algebra over $\mathbb{R}$ (necessarily $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$ by Frobenius's theorem), with dimension $1$, $2$, or $4$ over $\mathbb{R}$ rather than always $1$.
**Why is $\theta_0$ unique up to scalar?** This is the content of the case $V \cong W$. Any other $G$-isomorphism $\theta: V \to W$ differs from $\theta_0$ by composition with a $G$-automorphism of $V$ (or of $W$); by Schur 2, all such automorphisms are scalars. So $\theta = \lambda \theta_0$ for some $\lambda \in \mathbb{C}^\times$ (non-zero because $\theta$ is an isomorphism). Even non-zero $G$-homomorphisms (which are necessarily isomorphisms) are determined up to a non-zero scalar; including the zero homomorphism, the full space $\operatorname{Hom}_G(V,W)$ is $\mathbb{C}\theta_0$.
[/guided]
[/step]
