How can a representation be a direct sum of its subrepresentations?

Question

Reading the definition of Maschke's Theorem, I was very confused. It said that $U$ and $W$ are subrepresentations of a representation $V$, yet $V=W\oplus U$?

How is this possible?

Shouldn't $W\oplus U$ be a subrepresentation of a space with twice the dimension of $V$, since it will be pairs of elements of $V$?

Answer 1

No. If $\dim W=m$ and $\dim U=n$, then $\dim\left(W\bigoplus U\right)=m+n$.

Answer 2

This is a question about direct sums of vector spaces that is independent of whether or not the vector spaces are also group-representations.

It seems you define $W \oplus U$ as the space of pairs $(w, u)$ with $u \in U$ and $w \in W$. Now forget for a moment that $U$ and $W$ were obtained as subspaces of some other space $V$ and instead define $V$ as $W \oplus U$.

Then (in this slightly upside-down setting) your question seems to be:

'If we give the resulting big space $W \oplus U$ the name $V$, how can $U$ and $W$ be also be subspaces of $V$? Elements of $W \oplus U$ look like pairs of vectors, elements of $U$ look like 'ordinary' vectors. These are completely different types of object! What is going on here?'

What is going on is that we identify $U$ with the subspace of all elements of the form $(0, u)$ in $W \oplus U$ and identify $W$ with the subspace of all elements of the form $(w, 0)$. It is clear that $U$ and $\{(0, u) \colon u \in U\}$ are isomorphic and that the latter is a subspace of $V = W \oplus U$, hence it is 'allowed' to think about $U$ as a subspace of $V$ itself. (And similar for $W$.)

EDIT:

now in the other direction.

Theorem Let $V$ be a vector space and $U, W$ be subspaces satisfying

1. $U \cap W = \{0\}$
2. Every element $v \in V$ can be written in the form $w + u$ with $w \in W$ and $u \in U$.

Then $V$ is isomorphic to $W \oplus U$.

Proof: item 2 gives an easy way to create such an isomorphism: element $v$ is sent to the pair $(w, u)$ where $w \in W$, $u \in U$ are such that $v = w + u$. Item 2 says they exist. You may worry that they are not unique. They are unique, that is because of item 1. Say that $w' + u' = w + u$ then $w' - w = u - u'$. The left hand side lives in $W$ and the right hand side in $U$ so being equal, both live in $W \cap U$. Hence both are 0 by item 1. But $w' - w = 0$ implies $w' = w$ and similarly we get $u' = u$. So uniqueness and existence are both ok.

So there you have you isomorphism, the only question is why is it an isomorphism? You can check this formally, I will here just point out why it is the same isomorphism (or its inverse, depending on your perspective) as that from the first part of my answer.

The isomorphism of the second part sends elements of $U \subset V$ to elements of the form $(0, u) \in W \oplus U$ and elements of $W \subset V$ to elements of the form $(w, 0)$, so that's how you see that it is the same identification we were talking about last time.