Last lecture: linear combinations and spanning sets Let X = { k } be a set of vectors in a vector space V A linear combination of k is any vector of the form r + r + + r k k V for r + r + + r k k for scalars r r r k R The span of X = { k } is the set Span(X) = { r + r + + r k k : r r r k R } Span(X) is a vector subspace of V Span(X) is the smallest subspace of V containing the set X In fact the subspaces of V are all of the form Span(X) for some X V (The trick is to find a good choice of X) Subspace eample I: In this lecture we are going to look at eamples of vectors subspaces and spanning sets in order to consolidate the work we have done so far -
Recall that F = {f : R R} is the vector space of all functions from R to R Question Let F = { f F : f() = f() } Is F a subspace of F? To answer this we have to ask our three questions: Is F non-empty? This is OK because the zero function () F Is F closed under vector addition? Well if f g F then f() = f() and g() = g() so that (f + g)() = f() + g() = f() + g() = (f + g)() Hence f + g F so that F is closed under addition Is F closed under scalar multiplication? If f F and r R then f() = f() so that (rf)() = rf() = rf() = (rf)() So rf F and F is closed under multiplication Hence F is a vector subspace of F Now let s vary the last eample by a small amount Let F + = { f F : f() f() } Is F + a subspace of F? -
Once again we have to ask ourselves three questions: Is F + non-empty? This is OK because the zero function () F + as () = = () Is F + closed under vector addition? Well if f g F + then f() f() and g() g() so that (f + g)() = f() + g() f() + g() = (f + g)() Hence f + g F + so that F+ is closed under vector addition Is F + closed under scalar multiplication? If f F + and r R then f() f() and (rf)() = rf()? rf() = (rf)() This works only if r and it fails if r <!! Therefore rf / F + Consequently F + if r < is not a vector subspace of F Spanning sets for R m : In the last lecture we saw that: { } is a spanning set for R { } is a spanning set for R -
{ { { R m Moreover if a a m such that + a m } is a spanning set for R } is a spanning set for R } is a spanning set for a a a m + a m R m then there are a a a m unique = a + a Spanning sets for R The spanning set for R m on the last slide is the standard spanning set for R m In fact there is an infinite number of other spanning sets for R m Consider R The following sets are all spanning sets for R : - +
{ { { 6 { { ab } for any a b c R c } } 4 } 4 } Sets which do not span R We have to be careful however as it is also easy to write down sets of vectors which do not span R (and similarly for R m ) { } { } { } { } or or or { Span a b c R Span { ab c } = Span { } { = Span -4 } bc for any }
{ { 5 7 5 7 5 } 9 } + For the last two eamples note that: = and 5 7 5 9 5 7 = Polynomial vector spaces: Recall that P n = { p + p + p + + p n n : p p p p n R } is the vector space of polynomials of degree at most n Question Can we find a spanning set for P n? Let s start small and look at P = { p : p R } which is the vector space of constant (polynomial) functions Then P = Span() where is the constant polynomial () = Net P = { p + p : p p R } Therefore P = Span( ) P = { p + p + p : p p p R } -5
Therefore P = Span( ) The general case: P n = { p + p + p + + p n n : p p p n R } So P n = Span( n ) Note that any polynomial p() can be written in eactly one way as a linear combination of the elements of these spanning sets Spanning sets for polynomial vector spaces: Let p() = 5 and q() = + which we consider as elements of P Question When does f() = a + b + c P belong to Span(p q)? If f() Span(p q) then f() = sp()+tq() for some s t R That is a + b + c = s5 + t( + ) = (5s + t) + t This gives equations for s and t: 5s + t = a t = b = c -6
To solve these equations we use Gaussian elimination again Spanning sets for polynomial vector spaces: 5 a The augmented matri is b c Therefore these equations are consistent if and only if c = = Span(p q) { a + b : a b R } = P Now suppose that c = Then R = 5 R 5 5 a b c R = R 5 5 a b b 5 a R =R 5 R 5 a b b That is f() = ( 5 a b)p() + b q() Check: a + b = ( 5 a b)5 + b ( + ) Hence Span(p q) = P -7
Spanning sets for Null spaces: Let A = Question Find a spanning set for Null(A) We use Gaussian elimination on A: R =R R R = R R =R R w y z So the general solution to A = is s t = s = s + t t Therefore Null(A) = Span ( ) This is a two dimensional subspace of R 4 Describing the span of some { vectors: Question Let X = Describe Span(X) geometrically -8 }
Suppose that yz Span(X) Then we can write yz = s for some s t R + t = s+t s s+t To find s and t we have to solve the augmented matri equation: R y =R R y R z =R R z R =R R y z y Hence yz Span(X) if and only if y + z = This means that Span(X) is the plane + y z = in R { } That is Span(X) = y : + y z = z -9
The column space of a matri: Suppose that A is an n m matri Definition The column space of A is the vector subspace Col(A) of R n which is spanned by the columns of A That is if A = a a a m then Col(A) = Span ( ) a a a m We will see later that the column space and the null space of A are very closely related to each other Eample of the column space of a matri: Eample Let A = The vector for some a b c R 4 5 8 4 9 6 Describe Col(A) yz belong to Col(A) if and only if yz = a + b + c 4 5 9 4 8 6 That is we have to solve the system of equations a + b + 4c = a + 5b + 8c = y 4a + 9b + 6c = z -
Which leads to the augmented matri: 4 5 8 y 4 9 6 z 4 4 R 5 8 y =R R y R 4 9 6 z =R 4R z 4 4 R =R R y z y Therefore yz Col(A) if and only if + y z = That is Col(A) = { y z : + y z = } This is a plane in R Eample Let A = As before yz 7 5 describe Col(A) Col(A) if and only if it is a solution of the following augmented matri equation: -
7 y 5 z R =R R R =R R R =R +R y + z z R =R +R y + z z + y 7 Thus Col(A) = { y z y z : 7 y z = } which is a plane in R -