# cha6 - Chapter 6 EIGENVALUES AND EIGENVECTORS 6.1...

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Chapter 6EIGENVALUES ANDEIGENVECTORS6.1MotivationWe motivate the chapter on eigenvalues by discussing the equationax2+ 2hxy+by2=c,where not all ofa, h, bare zero. The expressionax2+ 2hxy+by2is calledaquadratic forminxandyand we have the identityax2+ 2hxy+by2=bracketleftbigxybracketrightbigbracketleftbiggahhbbracketrightbiggbracketleftbiggxybracketrightbigg=XtAX,whereX=bracketleftbiggxybracketrightbiggandA=bracketleftbiggahhbbracketrightbigg.Ais called the matrix of the quadraticform.We now rotate thex, yaxes anticlockwise throughθradians to newx1, y1axes. The equations describing the rotation of axes are derived asfollows:LetPhave coordinates (x, y) relative to thex, yaxes and coordinates(x1, y1) relative to thex1, y1axes. Then referring to Figure 6.1:115
116CHAPTER 6. EIGENVALUES AND EIGENVECTORSa45a27a54a63a0a0a0a0a0a0a0a0a18a64a64a64a64a64a64a64a64a73a0a0a0a0a0a0a0a0a18a64a64a64a64a64a64a64a64a82a12a12a12a12a12a12a12a12a12a12a64a64a64xyx1y1PQROαθFigure 6.1: Rotating the axes.x=OQ=OPcos (θ+α)=OP(cosθcosαsinθsinα)=(OPcosα)cosθ(OPsinα)sinθ=ORcosθPRsinθ=x1cosθy1sinθ.Similarlyy=x1sinθ+y1cosθ.We can combine these transformation equations into the single matrixequation:bracketleftbiggxybracketrightbigg=bracketleftbiggcosθsinθsinθcosθbracketrightbiggbracketleftbiggx1y1bracketrightbigg,orX=PY, whereX=bracketleftbiggxybracketrightbigg, Y=bracketleftbiggx1y1bracketrightbiggandP=bracketleftbiggcosθsinθsinθcosθbracketrightbigg.We note that the columns ofPgive the directions of the positivex1andy1axes. AlsoPis an orthogonal matrix – we havePPt=I2and soP-1=Pt.The matrixPhas the special property that detP= 1.A matrix of the typeP=bracketleftbiggcosθsinθsinθcosθbracketrightbiggis called arotationmatrix.We shall show soon that any 2×2 real orthogonal matrix with determinant
6.1. MOTIVATION117equal to 1 is a rotation matrix.We can also solve for the new coordinates in terms of the old ones:bracketleftbiggx1y1bracketrightbigg=Y=PtX=bracketleftbiggcosθsinθsinθcosθbracketrightbiggbracketleftbiggxybracketrightbigg,sox1=xcosθ+ysinθandy1=xsinθ+ycosθ. ThenXtAX= (PY)tA(PY) =Yt(PtAP)Y.Now suppose, as we later show, that it is possible to choose an angleθsothatPtAPis a diagonal matrix, say diag(λ1, λ2). ThenXtAX=bracketleftbigx1y1bracketrightbigbracketleftbiggλ100λ2bracketrightbiggbracketleftbiggx1y1bracketrightbigg=λ1x21+λ2y21(6.1)and relative to the new axes, the equationax2+ 2hxy+by2=cbecomesλ1x21+λ2y21=c, which is quite easy to sketch. This curve is symmetricalabout thex1andy1axes, withP1andP2, the respective columns ofP,giving the directions of the axes of symmetry.