MATH 304Linear AlgebraLecture 17:Basis and coordinates.Basis and dimensionDefinition. Let V be a vector space. A linearlyindependent spanning set for V is called a basis.Theorem Any vector space V has a bas is. If Vhas a finite basis, then all bases for V are finite andhave the s ame number of elements.Definition. The dimension of a vector space V ,denoted dim V , is the number of elements in any ofits bases.Examples. • dim Rn= n• Mm,n(R): the space of m×n m atricesdim Mm,n(R) = mn• Pn: polynomials of degree less than ndim Pn= n• P: the space o f all po lynomialsdim P = ∞• {0}: the trivial vector spacedim {0} = 0Basis and coordinatesIf {v1, v2, . . . , vn} is a basis for a vector space V ,then any vector v ∈ V has a unique representationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. The coefficients x1, x2, . . . , xnarecalled the coordinates of v with respect to theordered basis v1, v2, . . . , vn.The mappingvector v 7→ its coordinates (x1, x2, . . . , xn)is a one-to-one correspo ndence between V and Rn.This correspondence r es pects linear operations in Vand in Rn.Examples. • Coordinates of a vectorv = (x1, x2, . . . , xn) ∈ Rnrelative to the s tandardbasis e1= (1, 0, . . . , 0, 0), e2= (0, 1, . . . , 0, 0),. . . ,en= (0, 0, . . . , 0, 1) are (x1, x2, . . . , xn).• Coordinates of a matrixa bc d∈ M2,2(R)relative to the basis1 00 0,0 10 0,0 01 0,0 00 1are (a, b, c, d).• Coordinates of a polynomialp(x) = a0+ a1x + · · · + an−1xn−1∈ Pnrelative tothe basis 1, x, x2, . . . , xn−1are (a0, a1, . . . , an−1).Vectors u1=(2, 1) and u2=(3, 1) form a basis for R2.Problem 1. Find coordinates of the vectorv = (7, 4) with respect to the basis u1, u2.The desired coordinates x, y satisfyv = xu1+yu2⇐⇒2x + 3y = 7x + y = 4⇐⇒x = 5y = −1Problem 2. Find the vector w whose coordinateswith respect to the basis u1, u2are (7, 4).w = 7u1+ 4u2= 7(2, 1) + 4( 3, 1) = (26, 11)Change of coordinatesGiven a vector v ∈ R2, let (x, y ) be its standardcoordinates, i.e., coordinates with respect to thestandard basis e1= (1, 0), e2= (0, 1), and let(x′, y′) be its coordinates with respect to the basi sv1= (3, 1), v2= (2, 1).Problem. Find a relation between (x, y ) and (x′, y′).By definition, v = xe1+ ye2= x′v1+ y′v2.In standard coordinates,xy= x′31+ y′21=3 21 1x′y′=⇒x′y′=3 21 1−1xy=1 −2−1 3xyChange of coordinates in RnLet u1, u2, . . . , unbe a basis for Rn.Take any vector x ∈ Rn. Let (x1, x2, . . . , xn) be thestandard coordinates of x and (x′1, x′2, . . . , x′n) be thecoordinates of x with respect to the basisu1, u2, . . . , un.Problem 1. Given the standard coordinates(x1, x2, . . . , xn), find the nonstandard coordinates(x′1, x′2, . . . , x′n).Problem 2. Given the nonstandard coordinates(x′1, x′2, . . . , x′n), find the standard coordinates(x1, x2, . . . , xn).It turns out thatx1x2...xn=u11u12. . . u1nu21u22. . . u2n............un1un2. . . unnx′1x′2...x′n.The matrix U = (uij) does not depend on the v ector x.Columns of U are coordinates of vectorsu1, u2, . . . , unwith respect to the standard basis.U is called the transition matrix from the basisu1, u2, . . . , unto the standard basis e1, e2, . . . , en.This solves Problem 2. To solve Problem 1, we haveto use the inverse matrix U−1, which is thetransition matrix from e1, . . . , ento u1, . . . , un.Problem. Find the transition matrix from thestandard basis e1, e2, e3in R3to the basisu1= (1, 1, 0), u2= (0, 1, 1), u3= (1, 1, 1).The transition matrix from u1, u2, u3to e1, e2, e3isU =10 111 10 1 1.The transition matrix from e1, e2, e3to u1, u2, u3isthe inverse matrix A = U−1.The inverse matrix can be computed using rowreduction.(U | I ) =1 0 1 1 0 01 1 1 0 1 00 1 10 0 1Subtract the 1st row from the 2nd row:→1 0 11 0 00 1 0−1 1 00 1 10 0 1Subtract the 2nd row from the 3rd row:→1 0 1 1 0 00 1 0 −1 1 00 0 1 1 −1 1→1 0 1 1 0 00 1 0 −1 1 00 01 1 −1 1Subtract the 3rd row from the 1st row:→1 0 00 1 −10 1 0 −1 1 00 0 11 −1 1= (I | U−1)Thus A =0 1 −1−1 1 01 −1 1.The columns of A are coordinates of vectorse1, e2, e3with respect to the basis u1, u2, u3.Change of coordinates: general caseLet V be a vector space.Let v1, v2, . . . , vnbe a basis for V and g1: V → Rnbe thecoordinate mapping corresponding to this basis.Let u1, u2, . . . , unbe another basis for V and g2: V → Rnbe the coordinate mapping corresponding to this basis.Vg1ւg2ցRn−→ RnThe composition g2◦g−11is a transformation of Rn.It has the form x 7→ Ux, where U i s an n×n matrix.U is called the transition matrix from v1, v2. . . , vntou1, u2. . . , un. Columns of U are coordinates of the vectorsv1, v2, . . . , vnwith respect to the basis u1, u2, . . . , un.Problem. Find the transition matrix from thebasis p1(x) = 1, p2(x) = x + 1 , p3(x) = (x + 1)2tothe basis q1(x) = 1, q2(x) = x, q3(x) = x2for thevector space P3.We have to find coordinates of the po lynomialsp1, p2, p3with respect to the basis q1, q2, q3:p1(x) = 1 = q1(x),p2(x) = x + 1 = q1(x) + q2(x),p3(x) = (x+1)2= x2+2x+1 = q1(x)+2q2(x)+q3(x).Thus the transition matrix is1 1 10 1 20 0 1.Problem. Find the transition matrix from thebasis v1= (1, 2, 3), v2= (1, 0, 1), v3= (1, 2, 1) tothe basis u1= (1, 1, 0), u2= (0, 1, 1), u3= (1, 1, 1).To change coordinates from v1, v2, v3to u1, u2, u3,we first chang e coordinates from v1, v2, v3toe1, e2, e3, and then from e1, e2, e3to u1, u2, u3.Transition matrix from v1, v2, v3to e1, e2, e3:U1=1 1 12 0 23 1 1.Transition matrix from e1, e2, e3to u1, u2, u3:U2=1 0 11 1 10 1 1−1=0 1 −1−1 1 01 −1 1.Transition matrix from v1, v2, v3to u1, u2, u3:U2U1=0 1 −1−1 1 01 −1 11 1 12 0 23 1 1=−1 −1 11 −1 12 2
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