CHANGE OF BASIS FOR LINEAR SPACES

Transcription

1 CHANGE OF ASIS FOR LINEAR SPACES. Motivating Example Consider P 2 the space of polynomials of degree at most 2. Let : P 2 P 2 be defined by (p)(x) = p (x) p(x). his is a linear transformation. We would like to associate to a matrix (as this can make computations easier. o do so we need to pick a basis of P 2. Let be the standard basis, so We compute his gives = (, x, x 2 ) L (a + a x + a 2 x 2 ) = a. a 2 (a + a x + a 2 x 2 ) = (a + a x + a 2 x 2 ) (a + a x + a 2 x 2 ) a = a + 2a 2 x a a x a 2 x 2 = (a a ) + (2a 2 a )x a 2 x 2. p(x) = a + a x + a 2 x 2 (a a ) + (2a 2 a )x a 2 x 2 = (p)(x) L [p(x) = a a a 2 L a a 2a 2 a = [ (p)(x) a 2 Here : R 3 R 3 is a linear transformation. It has matrix [ = matrix of a linear transformation We generalize the preceding example. Let V be a linear space with basis = (v,..., v n ). For a linear transform : V V define a matrix, called the -matrix of by [ = [ [ (v ) [ (v n ) hat is, the columns of the matrix are precisely the -coordinate vectors of the images under of the elements of the basis. his is a direct generalization of the matrix of a linear transformation : R n R n with respect to a basis of R n.

2 2 CHANGE OF ASIS FOR LINEAR SPACES Let s write = [. his matrix ensures that the following diagrams hold V V p (p) L L L L R n R n [p [p = [ (p) where here the bottom arrow is multiplication by. In other words, to find the value of (p) one computes (p) = L ([p ) = L ([ [p ). EXAMPLE: Consider P 2 with basis = (, x, x 2 ) and (p) = p p we compute () = [ () = and (x) = x [ (x) = his is [ from before. and (x 2 ) = 2x x 2 [ (x 2 ) = [ = Image and Kernel 2 Fix a linear space V with basis = (v,..., v n ). he transformations L and L give a dictionary between V and Rn. In particular, they give a dictionary between the image and kernel of and of [. EXAMPLE: Consider the linear transformation : R 2 2 R 2 2 defined by [ [ (A) = A A. Determine a basis of ker( ) and Im ( ). Using the basis = (e, e 2, e 2, e 22 ) where [ [ [ [ e =, e 2 =, e 2 =, e 22 = One computes, = [ = [ [ (e ) [ (e 2 ) [ (e 2 ) [ (e 22 ) = 2 2 ([ ) 2 For example: [ (e 2 ) = = 2.

3 CHANGE OF ASIS FOR LINEAR SPACES 3 Moreover, 2 rref() = the pivot columns are st and 3rd and free columns are 2nd and 4th. he kernel of consists of solutions to x 2x 2 x 4 = and x 3 =. hat is, by taking x 2 =, x 4 = and x 2 =, x 4 = and solving these simple systems one obtains 2 ker() = ker(rref()) = span,. his corresponds to ker( ) = span ([ [ ) 2,. Likewise, the pivot columns form a basis of the image of and so Im () = span, 2. ([ [ ) Im ( ) = span(, Change of asis Matrix Fix a linear space V with bases = (v,..., v n ) and U = (w,..., w n ). Observe, that the map L : Rn R n is the composition of isomorphisms and hence is an isomorphism. We call S U = [ L Rn n the change of basis matrix from to U Clearly, this is an invertible matrix. We have the diagrams L L V R n p L (p) S U S U R n (p) = S U [p. EXAMPLE: Let V C be the space V = span(, cos(2x), sin(2x)). Compute S U for the bases (I will leave it to you to check these are both bases) = (, cos(2x), sin(2x)) U = (, cos 2 (x), sin(x) cos(x)) o do so observe, that a basic trigonometric identity tells us that cos(2x) = cos 2 (x) sin 2 (x) = 2 cos 2 (x) and sin(2x) = 2 cos(x) sin(x).

4 4 CHANGE OF ASIS FOR LINEAR SPACES In particular, L a b = (a + b cos(2x) + c sin(2x)) c = (a b + 2b cos 2 (x) + 2c cos(x) sin(x)) = a b 2b. 2c S U = Change of basis for subspaces EXAMPLE: Consider V to be the subspace of R 3 given by x + x 2 + x 3 =. his has bases 2 =, and U =,. We compute (L ( e )) = 2 = 2 + = 2 e + e 2 and Observe, (L ( e 2)) = = = e e 2 S U = [ 2 2 [ = 2 Here the left hand matrix has columns the elements of and the righthand side is the matrix with columns the elements of U. his last fact can be generalized as follows: heorem 5.. If V R n has basis = ( b,..., b m ) and U = ( u,..., u m ) then [ b bm = [ u u m S U.

5 CHANGE OF ASIS FOR LINEAR SPACES 5 6. Change of basis matrix and linear transformations Fix a linear space V with with dim(v ) = n. Suppose that : V V is linear transformation from V to V. If = (v,..., v n ) and U = (u,..., u n ) form two bases of V, then it is natural to ask what is the relationship between [ and [ U. hat is, what is the relationship between the matrix of with respect to the two bases. heorem 6.. In the above situation, if S = S U is the change of basis matrix, then [ U S = S[ and [ U = S[ S and [ = S [ S. A heuristic to remember the order of multiplication is the following: [ U eats a U-coordinate vector and so is multiplied on the right by S = S U (as this outputs U-coordinate vectors). his product eats -vectors and outputs -coordinate vectors. Similarly, [ outputs -coordinate vectors and so has to be multiplied on the left by S = S U (which eats -vectors). his product eats -coordinate vectors and outputs U-coordinate vectors. EXAMPLE: Consider V = span(, cos(2x), sin(2x)) C with basis = (, cos(2x), sin(2x)) and U = (, cos 2 (x), sin(x) cos(x)). One checks the map D : V V given by D(f) = f is a well defined linear transformation. Indeed, Clearly, D(a + b cos(2x)) + c sin(2x)) = 2b sin(2x) + 2c cos(2x) V. [D = 2. 2 As D() =, D(cos 2 (x)) = 2 cos(x) sin(x) and D(cos(x) sin(x)) = cos 2 (x) sin 2 (x) = + 2 cos 2 (x), [D U = 2 2 We check 2 [D U S U = 2 2 = S U [D = 2 2 = 4, these agree as expected.