# tat02 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Summer...

This preview shows pages 1–3. 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: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Summer School Math2061 2010 Tutorial 2 (week 2) — Solutions 1. For each of the sets of vectors X ⊂ R 3 below, explicitly describe all of the vectors in the subspace Span( X ) of R 3 . a ) X = { } . b ) X = n 1 1 1 o . c ) X = n 1 1 1 , 2 2 2 o . d ) X = n 1 1 1 , 1 o . e ) X = n 1 1 1 , 1 , 1 1 o . Solution Recall that if X = { v 1 , . . . , v k } then Span( X ) = { λ 1 v 1 + ··· + λ k v k | λ 1 , . . . , λ k ∈ R } . a ) Span( ) = n o . b ) Span 1 1 1 = n r r r r ∈ R o . c ) As 2 2 2 = 2 1 1 1 , or by part (b), Span 1 1 1 , 2 2 2 = n r r r r ∈ R o . d ) Span 1 1 1 , 1 = n r + s r r r, s ∈ R o . We can also describe Span( X ) = n x y z y = z o geometrically as the plane in R 3 with equation y- z = 0 . To see this observe that a vector x y z ∈ Span( X ) if and only if x y z = r 1 1 1 + s 1 , for some r, s ∈ R . Using Gaussian elimination we can simplify this to 1 1 x 1 y 1 z ! R 2 := R 1- R 2 R 3 := R 1- R 3-------→ 1 1 x 1 x- y 1 x- z ! R 3 := R 3- R 2-------→ 1 1 x 1 x- y y- z ! Hence, x y z ∈ Span( X ) if and only if y- z = 0 ; that is, if and only if y = z . Conse- quently, we have that Span( X ) = n r + s r r r, s ∈ R o = n r s s r, s ∈ R o e ) Span 1 1 1 , 1 , 1 1 = n r + s + t r + t r r, s, t ∈ R o . Again, we analyze Span( X ) alge- braically. A vector x y z ∈ Span( X ) if and only if x y z = a 1 1 1 + b 1 + c 1 1 , Summer School Math2061 Tutorial 2 (week 2) — Solutions Page 2 for some a, b, c ∈ R . Using Gaussian elimination we can simplify this to 1 1 1 x 1 1 y 1 z ! R 2 := R 1- R 2 R 3 := R 1- R 3-------→ 1 1 1 x 1 x- y 1 x- z ! Hence, c = x- z , b = x- y and a = x- ( x- y )- ( x- z ) =- x + y + z . In particular, for any vector x y z we can always find suitable a, b, c ∈ R , so Span( X ) = n r s t r, s, t ∈ R o = R 3 . Note that we now have given two quite different “explicit” descriptions of Span( X ) : n r + s + t r + t r r, s, t ∈ R o = Span( X ) = n r s t r, s, t ∈ R o . A priori , it is not obvious that the left and right hand sides of this equation give the same subspace of R 3 . 2. Recall that F is the vector space of functions from R to R , with the usual operations of addition and scalar multiplication of functions. Determine which of the following subsets of F are vector subspaces of F and, in each case, find two functions which are in the set....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

tat02 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Summer...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online