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1、Chapter6EIGENVALUESANDEIGENVECTORS6.1MotivationWemotivatethechapteroneigenvaluesbydiscussingtheequationax2+2hxy+by2=c,wherenotallofa,h,barezero.Theexpressionax2+2hxy+by2iscalledaquadraticforminxandyandwehavetheidentity22ahxtax+2hxy+by=xy=XAX,hbyxahwhereX=andA=.Aiscalledthematr
2、ixofthequadraticyhbform.Wenowrotatethex,yaxesanticlockwisethroughθradianstonewx1,y1axes.Theequationsdescribingtherotationofaxesarederivedasfollows:LetPhavecoordinates(x,y)relativetothex,yaxesandcoordinates(x1,y1)relativetothex1,y1axes.ThenreferringtoFigure6.1:115116CHAPTER6.EIGENVALUESA
3、NDEIGENVECTORSyP6 @y1@x1@I@ @ R@ @ @ @α @ @ θQ-xO@@@@@@@@R?Figure6.1:Rotatingtheaxes.x=OQ=OPcos(θ+α)=OP(cosθcosα−sinθsinα)=(OPcosα)cosθ−(OPsinα)sinθ=ORcosθ−PRsinθ=x1cosθ−y1sinθ.Similarlyy=x1sinθ+y1cosθ.Wecancombinethesetransformationequationsintothesinglematrixequation:xcosθ−si
4、nθx1=,ysinθcosθy1xx1cosθ−sinθorX=PY,whereX=,Y=andP=.yy1sinθcosθWenotethatthecolumnsofPgivethedirectionsofthepositivex1andy1axes.AlsoPisanorthogonalmatrix–wehavePPt=I2andsoP−1=Pt.ThematrixPhasthespecialpropertythatdetP=1.cosθ−sinθAmatrixofthetypeP=iscalledarotationmatrix.sinθcosθ
5、Weshallshowsoonthatany2×2realorthogonalmatrixwithdeterminant6.1.MOTIVATION117equalto1isarotationmatrix.Wecanalsosolveforthenewcoordinatesintermsoftheoldones:x1tcosθsinθx=Y=PX=,y1−sinθcosθysox1=xcosθ+ysinθandy1=−xsinθ+ycosθ.ThenXtAX=(PY)tA(PY)=Yt(PtAP)Y.Nowsuppose,aswelatershow,tha
6、titispossibletochooseanangleθsothatPtAPisadiagonalmatrix,saydiag(λ1,λ2).Thentλ10x122XAX=x1y1=λ1x1+λ2y1(6.1)0λ2y1andrelativetothenewaxes,theequationax2+2hxy+by2=cbecomesλ1x21+λ2y12=c,whichisquiteeasytosketch.Thiscurveissymmetricalaboutthex1andy1axes,withP1andP2,therespectivecolumns
7、ofP,givingthedirectionsoftheaxesofsymmetry.AlsoitcanbeverifiedthatP1andP2satisfytheequationsAP1=λ1P1andAP2=λ2P2.u1Theseequationsforcearestrictiononλ1andλ2.ForifP1=,thev1firstequationbecomesahu1u1a−λ1hu10=λ1or=.hbv1v1hb−λ1v10Hencewearedealingwithahomogeneoussystemoft
