Working with external and internal direct sums

**Runty** · October 6th 2010, 11:20 AM

My information for this one is limited, so I'll need a hand.

Suppose $V$ and $W$ are vector spaces over a field $F$ .

Now suppose $W_1$ and $W_2$ are subspaces of $V$ . We denote the external direct sum of $W_1$ and $W_2$ by $W_1\times W_2$ , and denote the internal direct sum of $W_1$ and $W_2$ by $W_1\oplus W_2$ . Define $\varphi : W_1\times W_2\rightarrow W_1+W_2$ by $\varphi(w_1,w_2)=w_1+w_2$ for all $(w_1,w_2)\in W_1\times W_2$ .

a) Prove that $\varphi$ is a surjective linear transformation.

b) Prove that $\varphi$ is an isomorphism if and only if $W_1+W_2=W_1\oplus W_2$ .

There isn't anything in our reading material that differentiates between external and internal direct sums, though I found the definitions online. In case anyone needs the definition for direct sum (in a basic sense), I have it below.

$V=W_1\oplus ...\oplus W_k$ , if $W_1+...+W_k=V$ and $W_1,...,W_k$ are independent.

**Bruno J.** · October 6th 2010, 01:18 PM

$W_1+W_2 = \{w_1+w_2 : w_1 \in W_1, w_2 \in W_2\}$ . It's a subspace of $V$ which contains both $W_1$ and $W_2$ (and in fact the smallest such subspace). In the event that every element of $W_2+W_2$ can be written uniquely as $w_1+w_2$ , then we write $W_2\oplus W_2$ instead.

Both are direct consequences of the definitions. It's very easy to see that $\varphi$ is linear. To show that $\varphi$ is surjective, take any $w_1+w_2 \in W_1+W_2$ ; then $\varphi(w_1,w_2)=w_1+w_2$ .

The second one isn't any harder! Give it a try and post your idea.

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