081st_tut6sol

# 081st_tut6sol - MATH1111/2008-09/Tutorial VI Solution 1...

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

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.

Unformatted text preview: MATH1111/2008-09/Tutorial VI Solution 1 Tutorial VI Suggested Solution 1. Let V be a vector space and dim V = 2008, and let S,T be subspaces of V . Prove or disprove the following. (a) For every integer r with 0 ≤ r ≤ 2008, 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 2008 } be a basis for V . (a) True. For r = 0, we take W = { } which is a subspace of V and dim W = 0. For 1 ≤ r ≤ 2008, 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 .) However the ”if” part is wrong. Consider V = R 2 and let S = Span( e 1 ) and T = Span( e 2 ). Both S and T are subspaces of V . And dim S = 1 = dim T but S 6 = T . (c) True. Note that S ∩ T is a subspace of V when S and T are subspaces of V . As S ∩ T ⊂ S , dim( S ∩ T ) ≤ dim S . (See lecture slides.) As S ∩ T ⊂ T , dim( S ∩ T ) ≤ dim T ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

081st_tut6sol - MATH1111/2008-09/Tutorial VI Solution 1...

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

View Full Document
Ask a homework question - tutors are online