This lecture: basis and dimension 4.4. Linear Independence: Suppose that V is a vector space and. r 1 x 1 + r 2 x r k x k = 0

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1 Linear Independence: Suppose that V is a vector space and that x, x 2,, x k belong to V {x, x 2,, x k } are linearly independent if r x + r 2 x r k x k = only for r = r 2 = = r k = The vectors x, x 2,, x k are linearly dependent if they are not linearly independent; that is, if there exist scalars r, r 2,, r k which are not all zero such that r x + r 2 x r k x k = A basis of V is a set of linearly independent vectors which span V This lecture: basis and dimension 44 Question Why is this useful? Example Is {cos x, sin x, } is linearly independent? If s cos x + t sin x + r = then -

2 x = : s + t + r = x = π 2 : s + t + r = x = π 4 : s 2 + t 2 + r = Therefore, {cos x, sin x, } is linearly independent The order of the logic is very important here: For any particular value x = a of x we can find r, s, t R such that r + s cos a + t sin a = The point is that we have to find r, s, t R such that r + s cos x + t sin x = for all x R If we pick good test values of x then we can show that we must have r = s = t = Basis of a Vector Space: We now combine spanning sets and linear independence Definition Suppose that V is a vector space A basis of V is a set of vectors {x, x 2,, x k } in V such that V = Span(x, x 2,, x k ) and {x, x 2,, x k } is linearly independent -

3 Examples { [ [, { [ [ [,, is a basis of R 2 } is a basis of R 3 { [ [ [ [ },,,, is a basis of R m {, x, x 2 } is a basis of P 2 {, x, x 2,, x n } is a basis of P n Typically, if W is a vector subspace of V then our challenge is to find a basis for W [ [,, { [ Another basis of R 3 From the last slide, is a basis of R 3 There are many other bases of R 3 { [ Example Show that X = 2 3 basis of R 3 We need to check two things: -2 [ [,, } } is another

4 R 3 = Span(X) X is linearly independent [ R 3 x = Span(X): Suppose that yz R 3 [ x Then yz Span(X) if and only if we can find r, s, t R such that [ [ [ [ x yz = r + s + t 2 3 We apply Gaussian elimination: [ x R 2 y 2 =R 2 2R R 3 z 3 =R 3 3R [ x R 2 = R 2 2x y 2 z 3x R =R R 2 R 3 =R 3 +2R 2 [ y x 2x y x 2y + z Therefore, [ [ [ x yz = (y x) 2 + (2x y) 3 [ x 2 y 3 z [ x y 2x 2 z 3x [ + (x 2y + z) -3

5 Hence, Span(X) = R 3 We also need to check that X is linearly independent [ [ x Taking yz = = above, [ [ [ we see that = + + is the only linear 2 3 combination of X giving the zero vector Hence, X is linearly independent Therefore, X is a basis of R 3 The independence theorem Suppose that x, x 2,, x d is a basis of V and let v V Then v can be expressed as a linear combination of {x, x 2,, x d } in exactly one way Proof Suppose that r x + r 2 x r d x d = v = s x + s 2 x s d x d, for some r, r 2,, r d, s, s 2,, s d R So = v v = (r x + r 2 x r d x d ) (s x + s 2 x s d x d ) = (r s )x + (r 2 s 2 )x (r d s d )x d -4

6 But, x, x 2,, x d are linearly independent so this means that r s =, r 2 s 2 =,, r d s d = That is, r = s, r 2 = s 2,, r d = s d Hence, we can write v as a linear combination of x, x 2,, x d in a unique way as claimed! How big can a basis be? Suppose that we could find a basis {w, x, [ y, z} of R 3 [ with four elements [ [ w x y z Write w = w 2, x = x 2, y = y 2 and z = z 2 w 3 x 3 y 3 z 3 Let a, b, c, d R be scalars such that aw + bx + cy + dz = [ [ [ w x y That is, a + b + c y 2 y 3 w 2 w 3 x 2 x 3 [ z + d z 2 z 3 To solve this we use Gaussian elimination: [ w x y z w 2 x 2 y 2 z 2 w 3 x 3 y 3 z 3 = [ [ (at best) We must have at least one free variable So there is no way that {w, x, y, z} can be linearly independent -5

7 The dependence theorem Suppose that {x, x 2,, x d } is basis of V Then every linearly independent subset of V has at most d elements Proof Let y, y 2,, y n are vectors in V, where n > d We have to show the vectors y, y 2,, y n are linearly dependent That is, we have to show that we can find scalars r, r 2,, r n which are not all zero and r y + r 2 y r n y n = As {x, x 2,, x d } is basis of V we can certainly write: y = a x + a 2 x a d x d y 2 = a 2 x + a 22 x a 2d x d y 3 = a 3 x + a 32 x a 3d x d y n = a n x + a n2 x a nd x d ( ) Hence, r a ( x + a 2 x a d x d ) +r 2 a2 x + a 22 x a 2d x d ( ) +r n an x + a n2 x a nd x d = -6

8 Rearranging the last equation we have: ( r a + r 2 a r n a n ) x + ( r a 2 + r 2 a r n a n2 ) x2 + ( ) r a d + r 2 a 2d + + r n a nd xd = However, x, x 2,, x d are linearly independent, so: r a + r 2 a r n a n = r a 2 + r 2 a r n a n2 = r a d + r 2 a 2d + + r n a nd = This is a system of d equations in the n unknowns r, r 2,, r n As n > d there are infinitely many solutions In particular, we must have a non zero solution to r y + r 2 y r n y n = So, {y, y 2,, y n } is linearly dependent, as claimed -7

9 Basis Theorem 2 Suppose that {x, x 2,, x d } is a basis of V and that {y, y 2,, y n } is a linearly independent subset of V By the last result we must have n d The dimension theorem Every basis of V has the same size That is, if {x, x 2,, x d } and {y, y 2,, y n } are two bases of V then n = d Proof As {x, x 2,, x d } is a basis of V and {y, y 2,, y n } is linearly independent we have n d Similarly, as {y, y 2,, y n } is a basis of V and {x, x 2,, x d } is linearly independent we have d n Hence, n d n So n = d! Definition Suppose that V is a vector space with basis {x, x 2,, x d } Then the dimension of V is dim V = d -8

10 Dimensions of common vector spaces Examples { [ [ dim R 2 = 2 since, is a basis of R 2 { [ dim R 3 = 3 since R 3 dim R m = m since is a basis of R m { [ [ [,,, [ } is a basis of,, [ dim P = since {} is a basis of P dim P = 2 since {, x} is a basis of P dim P 2 = 3 since {, x, x 2 } is a basis of P 2, [ dim P n = n + since {, x, x 2,, x n } is a basis of P n dim P = dim F = } -9

11 Example Let a(x) =, b(x) = x and c(x) = (x ) 2 Is {a(x), b(x), c(x)} a basis of P 2? Let p(x) = u + vx + wx 2 be an arbitrary element of P 2 Then p(x) Span ( a(x), b(x), c(x) ) if and only if u + vx + wx 2 = ra(x) + sb(x) + tc(x), for some r, s, t R That is, u + vx + wx 2 = r + s(x ) + t(x 2 2x + ) Equating coefficients we require: x : r s + t = u x : s 2t = v x 2 : t = w Hence, p(x) = (u+v +w)a(x)+(v +2w)b(x)+wc(x) Check: u + vx + wx 2 = (u + v + w) + (v + 2w)(x ) + w(x 2 2x + ) Therefore, Span(a(x), b(x), c(x)) = P 2 Question Does this mean that {a(x), b(x), c(x)} must be linearly independent? -