# Working with external and internal direct sums

• Oct 6th 2010, 11:20 AM
Runty
Working with external and internal direct sums
My information for this one is limited, so I'll need a hand.

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

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

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

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

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.

$\displaystyle V=W_1\oplus ...\oplus W_k$, if $\displaystyle W_1+...+W_k=V$ and $\displaystyle W_1,...,W_k$ are independent.
• Oct 6th 2010, 01:18 PM
Bruno J.
$\displaystyle W_1+W_2 = \{w_1+w_2 : w_1 \in W_1, w_2 \in W_2\}$. It's a subspace of $\displaystyle V$ which contains both $\displaystyle W_1$ and $\displaystyle W_2$ (and in fact the smallest such subspace). In the event that every element of $\displaystyle W_2+W_2$ can be written uniquely as $\displaystyle w_1+w_2$, then we write $\displaystyle W_2\oplus W_2$ instead.

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

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