{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_18- Subspaces

# lecture_18- Subspaces - MA 265 LECTURE NOTES WEDNESDAY...

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

MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 20 Subspaces Definition. Let ( V, , ) be a real vector space. We say that a subset W V is a subspace of V if (1) W negationslash = , and (2) ( W, , ) is a real vector space. We spend the rest of the lecture giving examples and discussing relevant results. Determining Subspaces. Let ( V, , ) be a real vector space, and W V be a subset. We give a method to determine when W is a subspace. We will show that W is a subspace of V if and only if (i) W is nonempty: W negationslash = . (ii) W is closed under : If u , v W then u v W . (iii) W is closed under : If c R and u W then c u W . Hence in order to check that a subset is a subspace, it suffices to check that W is nonempty, closed under vector addition, and closed under scalar multiplication. We explain why this is true. First assume that W is a subspace of V . Then ( W, , ) is a real vector space. This means u v W for all u , v W so that (ii) holds; and c u W for all c R and u W so that (iii) holds. Since 0 W this shows that (i) holds. Now assume that properties (i), (ii), and (iii) hold. We must show that (1) W negationslash = , and (2) ( W, , ) is a real vector space. That is, if u , v , w W and c, d R then we must show that the following properties are valid: ( Commutativity: ) u v = v u . ( Associativity :) u ( v w ) = ( u v ) w , and c ( d u ) = ( cd ) u . ( Distributivity: ) c ( u v ) = ( c u ) ( c v ), and ( c + d ) u = ( c u ) ( d u ).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

lecture_18- Subspaces - MA 265 LECTURE NOTES WEDNESDAY...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online