# Proving Schur's lemma

Schur's Lemma: Let $(\Pi_i,V_i)$, $i=1, 2$ be two irreducible representations of a group $G$, and let $\phi : V_1 \to V_2$ be an intertwiner. Then either $\phi = 0$ or $\phi$ is a vector space isomorphism.

To prove a part of this lemma, my textbook considers the kernel $K$ of $\phi$, which is a subspace of $V_1$. Then it turns out that $K$ is invariant under $\Pi_1(g)$ for $g\in G$, $$\phi(\Pi_1(g)v) = \Pi_2(g)(\phi(v)) = \Pi_2(g)(0) = 0$$ for $v\in K$. Therefore $\Pi_1(g)v\in K$, asserting that $K$ is invariant under $\Pi_1(g)$. Since it has been hypothesized that $(\Pi_1,V_1)$ is irreducible, $K$ can only equal to either $\{0\}$ or $V_1$ itself. If one chooses $K=V_1$, then $\phi=0$, proving half of the lemma. But this is my question, how does taking $K=V_1$ imply $\phi=0$?

• becouse then the kernel coincides with whole space – Leox Jun 14 '16 at 10:07
• If the kernel is everything, then the map is $0$, by definition of the kernel. Also note that this is not "really" Schur's lemma (though it is sometimes stated as such). Schur's lemma is the stronger statement one gets when working over an algebraically closed field (that the isomorphisms are scalars). – Tobias Kildetoft Jun 14 '16 at 10:07
• May be there are some theorems in linear algebra that I have forgotten, but can someone tell me that is there a theorem that says in a finite dimensional vector space, when $\phi v= 0$ for all $v \in \textrm{dom }\phi$, then $\phi=0$? – nougako Jun 14 '16 at 10:11
• That is the definition of $\phi = 0$. – Tobias Kildetoft Jun 14 '16 at 10:12
• I am going to put this as nicely as I can, but if your confusion with Schur's lemma is caused by these things, then you really need to study more linear algebra before you study group representations. Otherwise, you will end up spending way too much time on things that are supposed to be trivial. – Tobias Kildetoft Jun 14 '16 at 10:17