*This preview shows
pages
1–2. Sign up to
view the full content.*

This ** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **integer need not be an integer. The rational numbers also works, as does the natural numbers. 1 3. (3 points)Let V be a vector space and let U and W be subspaces of V . Dene U + W = { u + w V | u U, w W } . Show that U + W is a subspace of V . First we check that is in U + V . This follows from taking the element + in U + V , noting that is in both U and in V . Now to check the closure properties, we need two write down two arbitrary elements of U + V . We will write these as u 1 + v 1 and u 2 + v 2 . Also, take a scalar a R . Then ( u 1 + v 1 ) + ( u 2 + v 2 ) = ( u 1 + u 2 ) + ( v 1 + v 2 ) . The vector on the right is in U + V because U and V are closed under addition (so we have written it as something in U + something in V ). Next we have a ( u 1 + v 1 ) = a u 1 + a v 1 . This is in U + V because U and V are closed under scalar multiplica-tion. 2...

View Full
Document