Invariant subspace $U$ of direct sum $V\oplus V$ of irreducible representation $V$ is isomorphic to $V$ I am studying from Hall's book "Lie Groups, Lie Algebras, and Representations" and I'm stumped on the following question:

Suppose that $V$ is an irreducible finite-dimensional representation of a group or Lie algebra over $\mathbb{C}$, and consider the associated representation $V\oplus V$. Show that every nontrivial invariant subspace $U$ of $V\oplus V$ is isomorphic to $V$ and is of the form $$U=\{(\lambda_1v,\lambda_2v)|v\in V\}$$ for some constants $\lambda_1$ and $\lambda_2$, not both zero.

From the looks of it, I think I should be using Schur's lemma, which tells me that if $V$ is an irreducible complex representation of a group or Lie algebra and $\phi:V\to V$ is an intertwining map of $V$ with itself, then $\phi=\lambda I$ for some $\lambda\in\mathbb{C}$. An intertwining map is a linear map which commutes with the action. My understanding is that other authors just call this a homomorphism of representations.
However, I do not see how to connect the end result with Schur's lemma. Perhaps the projection maps will come in handy here? Any help is appreciated.
 A: Here's a more detailed answer. First assume that $U$ is non-trivial irreducible invariant subspace of $V \oplus V$. For $i: U \hookrightarrow V \oplus V$ the inclusion and $pr_1 , pr_2: V \oplus V \twoheadrightarrow V$ the projections on the first and second component, repectively, look at the maps $$\phi_j:=pr_j \circ i: U \rightarrow V.$$
By Schur's Lemma, $\phi_j$ is with zero or isomorphism of representation. We cannot have both $\phi_j =0$. So we know for sure that $U$ is isomorphic to $V$.
The missing part in Torsten's answer is to show that $U$ has the form given in the question $$U=\{(\lambda_1v,\lambda_2v)|v\in V\}$$

*

*If $\phi_1 = 0$, then we know $U$ is the second copy of $V$ in $V\oplus V$ so $U$ takes the form required in the question with $\lambda_1 = 0, \lambda_2 = 1$.

*Similar argument if $\phi_2 = 0$.

*If both $\phi_1$ and $\phi_2$ are non-zero, we can apply the third statement of Schur's Lemma (as in Hall's Theorem 4.29) (note here we used the fact that we are working over $\mathbb{C}$) and conclude that $\phi_1 = \lambda \phi_2$ for some $\lambda\in\mathbb{C}$. This gives us $U=\{(v,\lambda v)|v\in V\}$.

Now what happens in general if $U$ is not irreducible, non-zero and invariant? Then it must contain a non-zero, invariant, irreducible subspace (this is an earlier question in Hall, but it's also straightforward to prove). The only such latter subspaces are all isomorphic to $V$, so $\dim U > \dim V$.
Consider $U\cap V$ where $V$ here is the first copy in $V\oplus V$. By considering dimensions, we must have $U\cap V$ non-zero. But $U\cap V$ is invariant. Since $V$ is irreducible, we must have $U\cap V$ equals to the first copy of $V$. A same argument gives $U$ intersect second copy of $V$ equals the second copy of $V$. Therefore, $U$ contains both copies of $V$ and therefore $V\oplus V$. $U$ must be the whole space.
[Alternative to the last paragraph: By proposition 4.26, we can write $V\oplus V$ as $U\oplus W$ where $W$ is another invariant subspace. If $\dim U > \dim V$, then $\dim W < \dim V$ and that can only mean $W = 0$.]
A: A hint would be to write down the short exact sequence
$$0\rightarrow U \rightarrow V \oplus V \rightarrow W \rightarrow 0.$$
We now have an intertwining map (i.e. homomorphism of representations) $\phi: V \oplus V \rightarrow W$.
This map, of course, is a direct sum of two maps $\phi_1: V \rightarrow W$ and $\phi_2: V \rightarrow W$.
What does Shur's lemma say about $\phi_1$ and $\phi_2$?
A: Step 1:  Assume $U$ is irreducible. For $i: U \hookrightarrow V \oplus V$ the inclusion and $pr_1 , pr_2: V \oplus V \twoheadrightarrow V$ the projections on the first and second component, repectively, look at the maps $$pr_j \circ i: U \rightarrow V.$$ Use Schur's Lemma. Notice what happens if both are zero.
Step 2: Now, assume $U$ contains two different irreducible subrepresentations $U_1 \neq U_2$. Using the result of step 1, show that $U_1+U_2$ contains both $V \oplus 0$ and $0 \oplus V$, hence is all of $V \oplus V$.
