082nd_tut5sol

082nd_tut5sol - MATH1111/2008-09/Tutorial V Solution 1...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH1111/2008-09/Tutorial V Solution 1 Tutorial V Suggested Solution 1. Let V be a vector space and dim V = 2009, and let S,T be subspaces of V . Prove or disprove the following. (a) For every integer r with 0 ≤ r ≤ 2009, one can always find a subspace W r of V such that dim W r = r . (b) S = T if and only if dim S = dim T . (c) dim( S ∩ T ) ≤ min(dim S, dim T ) where min( a,b ) = the minimum of a and b . (d) dim( S ∪ T ) ≥ max(dim S, dim T ) where max( a,b ) = the maximum of a and b . (e) Let v ∈ V and U = Span( v ). Then dim( U + S ) = dim S + 1. Ans . Let { v 1 ,v 2 , ··· ,v 2009 } be a basis for V (since dim V = 2009). (a) True. For r = 0, we take W = { } which is a subspace of V and dim W = 0. For 1 ≤ r ≤ 2009, we take W r to be the subspace Span( v 1 ,v 2 , ··· ,v r ). You can check easily that v 1 ,v 2 , ··· ,v r forms a basis for W r . So dim W r = r . (b) False. The ”only if” part is obviously correct (i.e. S = T are subspaces ⇒ dim S = dim T .)....
View Full Document

This note was uploaded on 12/06/2010 for the course MATH MATH101 taught by Professor Chan during the Spring '09 term at HKUST.

Page1 / 4

082nd_tut5sol - MATH1111/2008-09/Tutorial V Solution 1...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online