[proofplan] Both parts rest on the same observation: the kernel and image of a $G$-homomorphism $\theta$ are $G$-subspaces of the source and target respectively. Combined with the irreducibility hypotheses, this forces $\theta$ to be either zero or an isomorphism (part 1). For part 2, take an eigenvalue $\lambda \in \mathbb{F}$ — guaranteed to exist since $\mathbb{F}$ is algebraically closed — and apply part 1 to $\theta - \lambda \iota_V$, whose kernel contains the corresponding eigenvector and so cannot be the zero subspace. Part 1 then forces $\theta - \lambda \iota_V = 0$, giving $\theta = \lambda \iota_V$. [/proofplan] [step:Verify that $\ker\theta$ and $\operatorname{im}\theta$ are $G$-subspaces (proof of part 1)] Let $\theta: V \to W$ be a $G$-homomorphism, meaning an $\mathbb{F}$-linear map satisfying $\theta(gv) = g\theta(v)$ for all $g \in G$ and $v \in V$. The kernel $\ker\theta = \{v \in V : \theta(v) = 0\}$ is an $\mathbb{F}$-subspace of $V$ since $\theta$ is $\mathbb{F}$-linear. It is $G$-invariant: for $v \in \ker\theta$ and $g \in G$, \begin{align*} \theta(gv) = g\theta(v) = g \cdot 0 = 0, \end{align*} so $gv \in \ker\theta$. Hence $\ker\theta$ is a $G$-subspace of $V$. The image $\operatorname{im}\theta = \{\theta(v) : v \in V\}$ is an $\mathbb{F}$-subspace of $W$ since $\theta$ is $\mathbb{F}$-linear. It is $G$-invariant: for $w = \theta(v) \in \operatorname{im}\theta$ and $g \in G$, \begin{align*} gw = g\theta(v) = \theta(gv) \in \operatorname{im}\theta. \end{align*} Hence $\operatorname{im}\theta$ is a $G$-subspace of $W$. [/step] [step:Use irreducibility of $V$ and $W$ to constrain $\ker\theta$ and $\operatorname{im}\theta$ (proof of part 1)] Since $V$ is irreducible, its only $G$-subspaces are $0$ and $V$. Since $\ker\theta$ is a $G$-subspace of $V$, either $\ker\theta = 0$ or $\ker\theta = V$. Since $W$ is irreducible, its only $G$-subspaces are $0$ and $W$. Since $\operatorname{im}\theta$ is a $G$-subspace of $W$, either $\operatorname{im}\theta = 0$ or $\operatorname{im}\theta = W$. Two cases: - **$\ker\theta = V$.** Then $\theta(v) = 0$ for every $v \in V$, so $\theta = 0$. - **$\ker\theta = 0$.** Then $\theta$ is injective. We claim $\operatorname{im}\theta \neq 0$ in this case: since $V$ is irreducible, $V \neq 0$, so there exists $v_0 \in V$ with $v_0 \neq 0$, and $\theta(v_0) \neq 0$ by injectivity (as $\ker\theta = 0$). Thus $\operatorname{im}\theta \neq 0$, forcing $\operatorname{im}\theta = W$ by irreducibility of $W$. So $\theta$ is bijective, hence an isomorphism of $\mathbb{F}$-vector spaces; combined with the $G$-equivariance $\theta(gv) = g\theta(v)$, $\theta$ is a $G$-isomorphism. This concludes the proof of part 1. [guided] The two pieces of part 1 — that $\ker\theta$ and $\operatorname{im}\theta$ are $G$-subspaces — are direct from the definitions, but worth making explicit. **Why is $\ker\theta$ $G$-invariant?** A $G$-homomorphism satisfies $\theta(gv) = g\theta(v)$. Plugging in $v \in \ker\theta$, the right side is $g \cdot 0 = 0$, so the left side is also zero, meaning $gv \in \ker\theta$. The kernel is $G$-stable. **Why is $\operatorname{im}\theta$ $G$-invariant?** Take $w \in \operatorname{im}\theta$, so $w = \theta(v)$ for some $v$. Then $gw = g\theta(v) = \theta(gv)$, which is in $\operatorname{im}\theta$ since $gv \in V$. The image is $G$-stable. **Now the irreducibility squeeze.** Irreducibility of $V$ says only two $G$-subspaces exist in $V$: the zero subspace and $V$ itself. Same for $W$. So $\ker\theta \in \{0, V\}$ and $\operatorname{im}\theta \in \{0, W\}$. The four logical combinations: - $\ker\theta = V$, $\operatorname{im}\theta = 0$: $\theta = 0$. Possible. - $\ker\theta = V$, $\operatorname{im}\theta = W$: incompatible — $\ker\theta = V$ means $\theta$ is the zero map, but $\operatorname{im}\theta = W \neq 0$ requires non-zero outputs. Impossible. - $\ker\theta = 0$, $\operatorname{im}\theta = 0$: $\theta = 0$ (since the image is zero), but $\ker\theta = 0$ is forced, requiring $V = 0$, contradicting irreducibility ($V$ is non-zero by convention). Impossible. - $\ker\theta = 0$, $\operatorname{im}\theta = W$: $\theta$ is injective and surjective, hence an isomorphism. So either $\theta = 0$ or $\theta$ is a $G$-isomorphism. [/guided] [/step] [step:Apply algebraic closedness to extract an eigenvalue (proof of part 2)] Now suppose $\mathbb{F}$ is algebraically closed, $V$ is an irreducible $G$-space, and $\theta: V \to V$ is a $G$-endomorphism. Since $V$ is irreducible, $V$ is a non-zero finite-dimensional $\mathbb{F}$-vector space (irreducibility presumes finite-dimensionality in this context). The $\mathbb{F}$-linear endomorphism $\theta: V \to V$ has a characteristic polynomial $p_\theta(x) \in \mathbb{F}[x]$ of degree $\dim_\mathbb{F} V \geq 1$. Since $\mathbb{F}$ is algebraically closed, $p_\theta$ has at least one root $\lambda \in \mathbb{F}$, which is by definition an eigenvalue of $\theta$. [/step] [step:Apply part 1 to $\theta - \lambda \iota_V$ (proof of part 2)] Define \begin{align*} \theta - \lambda \iota_V: V \to V, \qquad v \mapsto \theta(v) - \lambda v. \end{align*} This is a $G$-endomorphism of $V$: - It is $\mathbb{F}$-linear, being a linear combination of $\mathbb{F}$-linear maps. - It is $G$-equivariant: for $g \in G$ and $v \in V$, \begin{align*} (\theta - \lambda \iota_V)(gv) = \theta(gv) - \lambda(gv) = g\theta(v) - g(\lambda v) = g(\theta(v) - \lambda v) = g \cdot (\theta - \lambda \iota_V)(v), \end{align*} using $G$-equivariance of $\theta$ and $\mathbb{F}$-linearity of the action of $g$. Since $\lambda$ is an eigenvalue of $\theta$, there exists a non-zero $v_\lambda \in V$ with $\theta(v_\lambda) = \lambda v_\lambda$, i.e., $(\theta - \lambda \iota_V)(v_\lambda) = 0$. So $\ker(\theta - \lambda \iota_V) \neq 0$, hence $\theta - \lambda \iota_V$ is **not** an isomorphism (a bijective linear map has zero kernel). By part 1 of Schur's lemma applied with $V = W$ to the $G$-endomorphism $\theta - \lambda \iota_V$ — which we just showed is not an isomorphism — we conclude $\theta - \lambda \iota_V = 0$. Equivalently, \begin{align*} \theta = \lambda \iota_V, \end{align*} completing the proof of part 2. [guided] **Why is part 2 a corollary of part 1, and not a separate argument?** Part 1 dichotomises $G$-homomorphisms between irreducibles into "zero or isomorphism." Part 2 says: among $G$-endomorphisms of an irreducible, every endomorphism takes the form $\lambda \iota_V$ for some scalar $\lambda$. The link is that $\theta - \lambda \iota_V$ is itself a $G$-endomorphism, and producing a $\lambda$ for which $\theta - \lambda \iota_V$ has a non-zero kernel is exactly producing a $\lambda$ for which $\theta - \lambda \iota_V$ falls into the "zero" branch of the dichotomy. **Why is algebraic closedness needed?** To produce the eigenvalue $\lambda \in \mathbb{F}$. Without algebraic closedness, the characteristic polynomial of $\theta$ might have no root in $\mathbb{F}$, and there would be no candidate scalar $\lambda$ for which $\theta - \lambda \iota_V$ has a non-zero kernel. The conclusion of part 2 fails in this case: e.g., over $\mathbb{R}$, the rotation by $90^\circ$ on $\mathbb{R}^2$ is an endomorphism of an irreducible representation of the cyclic group of order $4$, but it is not a real scalar multiple of the identity. **The role of finite-dimensionality.** The characteristic polynomial of $\theta$ has degree $\dim V$, which we need to be $\geq 1$ to guarantee a root. The convention in this course is that "irreducible representation" already implies finite-dimensional, so $\dim V \geq 1$. In infinite-dimensional settings (e.g., unitary representations of locally compact groups), part 2 still holds in the form $\theta = \lambda \iota_V$ but requires spectral theory rather than the characteristic polynomial. **Why does $\ker(\theta - \lambda\iota_V) \neq 0$ imply $\theta - \lambda \iota_V$ is non-invertible?** Because invertible linear maps are injective, and an injective linear map has zero kernel. The contrapositive: a non-zero kernel implies non-injective implies non-invertible. **Where does part 1 force $\theta - \lambda\iota_V = 0$?** Part 1 applied to a $G$-endomorphism (the case $V = W$) gives: a $G$-endomorphism of an irreducible is either zero or an isomorphism. We have established that $\theta - \lambda \iota_V$ is not an isomorphism, so it must be zero. [/guided] [/step]