{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 2.8 - 2.8 that Subspaces of Rn Denition A subspace of Rn is...

This preview shows pages 1–3. Sign up to view the full content.

2.8 Subspaces of R n Definition: A subspace of R n is a subset H∈ R n such that 1. vector 0 ∈H 2. For all vectoru , vectorv ∈H , vectoru + vectorv ∈H Closed under addition 3. For all vectoru ∈H , c R , cvectoru ∈H Closed under multiplication e.g. If vectorv 1 , ··· ,vectorv m R n , then H = Span { vectorv 1 , ··· vectorv m } is a subspace because 1. vector 0 ∈H as vector 0 = 0 · vectorv 1 + ··· + 0 · vectorv m . 2. vectoru = c 1 · vectorv 1 + ··· + c m · vectorv m vectorv = d 1 · vectorv 1 + ··· + d m · vectorv m = vectoru + vectorv = ( c 1 + d 1 ) · vectorv 1 + ··· + ( c m + d m ) · vectorv m ∈H 3. cvectoru = cc 1 · vectorv 1 + ··· + cc m · vectorv m ∈H H is the subspace spanned by vectorv 1 , ··· ,vectorv m . 1

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

View Full Document
EX: H = { ( x, y ) : x + y = 1 }⊂ R 2 . Is it a subspace? x + y = 1 y x What are subspaces of R 2 ? Recall that if vectorv 1 ,vectorv 2 R 2 are linearly independent, any vector vectorv R 2 is a linear combination of vectorv 1 and vectorv 2 , so Span { vectorv 1 ,vectorv 2 } = R 2 . If vectorv 1 = αvectorv 2 , Span { vectorv 1 ,vectorv 2 } = { tvector v 2 : t R } So Span { vectorv 1 ,vectorv 2 } is the straight line through the origin in the direction of vectorv 2 . •{ vector 0 } is a subspace.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern