MATH 304Linear AlgebraLecture 18:Basis and coordinates.Change of 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 (called thedimension of V ).Example. Vectors e1= (1, 0, 0, . . . , 0, 0),e2= (0, 1, 0, . . . , 0, 0),. . . , en= (0, 0, 0, . . . , 0, 1)form a basis for Rn(called standard) since(x1, x2, . . . , xn) = x1e1+ x2e2+ · · · + xnen.Basis 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 coeffici ents 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 matri xa bc d∈ M2,2(R)relative to the bas is1 00 0,0 01 0,0 10 0,0 00 1are (a, c, b, 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=(3, 1) and u2=(2, 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⇐⇒3x + 2y = 7x + y = 4⇐⇒x = −1y = 5Problem 2. Find the vector w whose coordinateswith respect to the basis u1, u2are (7, 4).w = 7u1+ 4u2= 7(3, 1) + 4(2, 1) = (29, 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 su1= (3, 1), u2= (2, 1).Problem. Find a relation between (x, y ) and (x′, y′).By definition, v = xe1+ y e2= x′u1+ y′u2.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 RnThe usual (standard) coordinates of a vectorv = (x1, x2, . . . , xn) ∈ Rnare coordinates relative to thestandard basis e1= (1, 0, . . . , 0, 0), e2= (0, 1, . . . , 0, 0),. . . ,en= (0, 0, . . . , 0, 1).Let u1, u2, . . . , unbe another basis for Rnand (x′1, x′2, . . . , x′n)be the coordinates of the same vector v with respect to thisbasis.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 v.Columns of U are coordinates of v ectorsu1, u2, . . . , unwith respect to the standard basis.U is cal led the transition matrix from the basisu1, u2, . . . , unto the standard basis e1, e2, . . . , en.This solves Problem 2. To s olve Problem 1, we haveto use the inverse matrix U−1, which is thetransition matrix from e1, . . . , ento u1, . . . , un.Problem. Find coordinates of the vectorx = (1, 2, 3) with respect to the basisu1= (1, 1, 0) , u2= (0, 1, 1) , u3= (1, 1, 1) .The nonstandard coordinates (x′, y′, z′) of x satisfyx′y′z′= U123,where U is the transition matrix from the standard basise1, e2, e3to the basis u1, u2, u3.The transition matrix from u1, u2, u3to e1, e2, e3isU0= (u1, u2, u3) =10 11 1 10 1 1.The transition matrix from e1, e2, e3to u1, u2, u3is theinverse matrix: U = U−10.The inverse matrix can be computed using row reduction.(U0| I ) =1 0 11 0 01 1 10 1 00 1 10 0 1→1 0 1 1 0 00 1 0−1 1 00 1 10 0 1→1 0 11 0 00 1 0−1 1 00 0 11 −1 1→1 0 00 1 −10 1 0−1 1 00 0 11 −1 1= (I | U−10)Thusx′y′z′=0 1 −1−1 1 01 −1 1123=−112.Change of coordinates: general caseLet V be a vector space of dimension n.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 is 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)2to the basis q1(x) = 1, q2(x) = x, q3(x) = x2forthe vector 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).Hence the transition matrix is1 1 10 1 20 0 1.Thus the polyno mial identitya1+ a2(x + 1) + a3(x + 1)2= b1+ b2x + b3x2is equivalent to the relationb1b2b3=1 1 10 1 20 0 1a1a2a3.Problem. Find the transition matrix from thebasis v1= (1, 2, 3) , v2= (1, 0, 1) , v3= (1, 2, 1) tothe basi s u1= (1, 1, 0), u2= (0, 1, 1), u3= (1, 1, 1) .It is convenient to make a two-step transition:first from v1, v2, v3to e1, e2, e3, and then f rome1, e2, e3to u1, u2, u3.Let U1be the transitio n matrix from v1, v2, v3toe1, e2, e3and U2be the transitio n matrix fromu1, u2, u3to e1, e2, e3:U1=1 1 12 0 23 1 1, U2=1 0 11 1 10 1 1.Basis v1, v2, v3=⇒ coordinates xBasis e1, e2, e3=⇒ coordinates U1xBasis u1, u2, u3=⇒ coordinates U−12(U1x)=(U−12U1)xThus the transition matrix from v1, v2, v3tou1, u2, u3is U−12U1.U−12U1=1 0 11 1 10 1 1−11 1 12 0 23 1 1=0 1 −1−1 1 01 −1 11 1 12 0 23 1 1=−1 …
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