Transcription of MATH 304 Linear Algebra
1 math 304 Linear AlgebraLecture 14:Basis and of and a vector space. A linearlyindependent spanning set forVis called vector spaceVhas a basis. IfVhas a finite basis, then all bases forVare finite andhave the same number of elements (called thedimensionofV). (1,0,0, .. ,0,0),e2= (0,1,0, .. ,0,0),.. ,en= (0,0,0, .. ,0,1)form a basis forRn(calledstandard) since(x1,x2, .. ,xn) =x1e1+x2e2+ + and coordinatesIf{v1,v2, .. ,vn}is a basis for a vector spaceV,then any vectorv Vhas a unique representationv=x1v1+x2v2+ +xnvn,wherexi R. The coefficientsx1,x2, .. ,xnarecalled thecoordinatesofvwith respect to theordered basisv1,v2, .. , mappingvectorv7 its coordinates(x1,x2, .. ,xn)is a one-to-one correspondence correspondence respects Linear operations inVand Coordinates of a vectorv= (x1,x2, .. ,xn) Rnrelative to the standardbasise1= (1,0, .. ,0,0),e2= (0,1.)
2 ,0,0),.. ,en= (0,0, .. ,0,1) are (x1,x2, .. ,xn). Coordinates of a matrix a bc d M2,2(R)relative to the basis 1 00 0 , 0 01 0 , 0 10 0 , 0 00 1 are (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).Vectorsu1=(2,1) andu2=(3,1) form a basis coordinates of the vectorv= (7,4) with respect to the basisu1, desired coordinatesx,ysatisfyv=xu1+yu2 2x+ 3y= 7x+y= 4 x= 5y= 1 Problem the vectorwwhose coordinateswith respect to the basisu1,u2are (7,4).w= 7u1+ 4u2= 7(2,1) + 4(3,1) = (26,11)Change of coordinatesGiven a vectorv R2, let (x,y) be its standardcoordinates, , coordinates with respect to thestandard basise1= (1,0),e2= (0,1), and let(x ,y ) be its coordinates with respect to the basisu1= (3,1),u2= (2,1). a relation between (x,y) and (x ,y ).By definition,v=xe1+ye2=x u1+y standard coordinates, xy =x 31 +y 21 = 3 21 1 x y = x y = 3 21 1 1 xy = 1 2 1 3 xy Change of coordinates inRnThe usual (standard) coordinates of a vectorv= (x1,x2.
3 ,xn) Rnare coordinates relative to thestandard basise1= (1,0, .. ,0,0),e2= (0,1, .. ,0,0),.. ,en= (0,0, .. ,0,1).Letu1,u2, .. ,unbe another basis forRnand (x 1,x 2, .. ,x n)be the coordinates of the same vectorvwith respect to the standard coordinates(x1,x2, .. ,xn), find the nonstandard coordinates(x 1,x 2, .. ,x n).Problem the nonstandard coordinates(x 1,x 2, .. ,x n), find the standard coordinates(x1,x2, .. ,xn).It turns out that = u11u12..u1nu21u22..unn x 1x n .The matrixU= (uij) does not depend on the ofUare coordinates of vectorsu1,u2, .. ,unwith respect to the standard called thetransition matrixfrom the basisu1,u2, .. ,unto the standard basise1,e2, .. , solves Problem 2. To solve Problem 1, we haveto use the inverse matrixU 1, which is thetransition matrix frome1, .. ,entou1, .. , coordinates of the vectorx= (1,2,3) with respect to the basisu1= (1,1,0),u2= (0,1,1),u3= (1,1,1).
4 The nonstandard coordinates (x ,y ,z ) ofxsatisfy x y z =U 123 ,whereUis the transition matrix from the standard basise1,e2,e3to the basisu1,u2, transition matrix fromu1,u2,u3toe1,e2,e3isU0= (u1,u2,u3) = 101111011 .The transition matrix frome1,e2,e3tou1,u2,u3is theinverse matrix:U=U 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 11 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)Thus x y z = 0 1 1 1 1 01 1 1 123 = 112 .Change of coordinates: general caseLetVbe a vector space of ,v2, .. ,vnbe a basis forVandg1:V Rnbe thecoordinate mapping corresponding to this ,u2, .. ,unbe another basis forVandg2:V Rnbe the coordinate mapping corresponding to this g2 Rn RnThe compositiong2 g 11is a transformation has the formx7 Ux, whereUis ann called thetransition matrixfromv1,v2.
5 ,vntou1,u2.. ,un. Columns ofUare coordinates of the vectorsv1,v2, .. ,vnwith respect to the basisu1,u2, .. , the transition matrix from thebasisp1(x) = 1,p2(x) =x+ 1,p3(x) = (x+ 1)2to the basisq1(x) = 1,q2(x) =x,q3(x) =x2forthe vector have to find coordinates of the polynomialsp1,p2,p3with respect to the basisq1,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 is 1 1 10 1 20 0 1 .Thus the polynomial identitya1+a2(x+ 1) +a3(x+ 1)2=b1+b2x+b3x2is equivalent to the relation b1b2b3 = 1 1 10 1 20 0 1 a1a2a3 . the transition matrix from thebasisv1= (1,2,3),v2= (1,0,1),v3= (1,2,1) tothe basisu1= (1,1,0),u2= (0,1,1),u3= (1,1,1).It is convenient to make a two-step transition:first fromv1,v2,v3toe1,e2,e3, and then frome1,e2,e3tou1,u2, the transition matrix fromv1,v2,v3toe1,e2,e3andU2be the transition matrix fromu1,u2,u3toe1,e2,e3:U1= 1 1 12 0 23 1 1 ,U2= 1 0 11 1 10 1 1.
6 Basisv1,v2,v3= coordinatesxBasise1,e2,e3= coordinatesU1xBasisu1,u2,u3= coordinatesU 12(U1x)=(U 12U1)xThus the transition matrix fromv1,v2,v3tou1,u2,u3isU 12U1= 1 0 11 1 10 1 1 1 1 1 12 0 23 1 1 = 0 1 1 1 1 01 1 1 1 1 12 0 23 1 1 = 1 1 11 1 12 2 0 . Linear mapping= Linear transformation= Linear vector spacesV1andV2, amappingL:V1 V2islinearifL(x+y) =L(x) +L(y),L(rx) =rL(x)for anyx,y V1andr Linear mapping :V Ris called (or if bothV1andV2are functionalspaces) then a Linear mappingL:V1 V2is calledalinear mapping= Linear transformation= Linear vector spacesV1andV2, amappingL:V1 V2islinearifL(x+y) =L(x) +L(y),L(rx) =rL(x)for anyx,y V1andr functionf:R Rgiven byf(x) =ax+bis a Linear transformation of thevector spaceRif and only ifb= of Linear mappings Scaling L:V V,L(v) =sv, wheres (x+y) =s(x+y) =sx+sy=L(x) +L(y),L(rx) =s(rx) =r(sx) =rL(x).
7 Dot product with a fixed vector :Rn R, (v) =v v0, wherev0 Rn. (x+y) = (x+y) v0=x v0+y v0= (x) + (y), (rx) = (rx) v0=r(x v0) =r (x). Cross product with a fixed vectorL:R3 R3,L(v) =v v0, wherev0 R3. Multiplication by a fixed matrixL:Rn Rm,L(v) =Av, whereAis anm nmatrix and all vectors are column mappings of functional vector spaces Evaluation at a fixed point :F(R) R, (f) =f(a), wherea R. Multiplication by a fixed functionL:F(R) F(R),L(f) =gf, whereg F(R). Differentiation D:C1(R) C(R),L(f) =f .D(f+g) = (f+g) =f +g =D(f) +D(g),D(rf) = (rf) =rf =rD(f). Integration over a finite interval :C(R) R, (f) =Zbaf(x)dx, wherea,b R,a<b.
