# A corresponding to eigenvalues ? 1 2 λ2 3 find

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Acorresponding to eigenvaluesλ1= 2,λ2=3,find matricesDandPin an orthogonal diag-onalization ofA.1.D=2003,P=21122.D=2003,P=1521123.D=3002,P=15122
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Version 047 – NewEXAM04 – gilbert – (53525)34.D=2003,P=151221correct5.D=3002,P=12216.D=2003,P=1221Explanation:WhenD=λ100λ,Q= [u1u2],thenQhas orthogonal columns andA=QDQ1is a diagonalization ofA, but it isnot an orthogonal diagonalization becauseQis not an orthogonal matrix. We have to nor-malizeu1andu2: setv1=u1u1=1512,v2=u2u2=1521.ThenP= [v1v2] is an orthogonal matrixandA=PDP1is an orthogonal diagonalization ofAwhenD=2003,P=151221.00510.0 pointsAnn×nmatrix that is orthogonally diag-onalizable must be symmetric.True or False?1.TRUEcorrect2.FALSEExplanation:By definition, ifAis orthogonally diagonal-izable, thenA=PDP1=PDPT,where D is a diagonal matrix and P is anorthogonal matrix, that isPT=P1. ThusAT= (PDPT)T= (PT)TDTPT=PDPT=Abecause (AB)T=BTATfor all matricesA, BandDT=Dfor any diagonal matrix. ThusA=AT, soAis symmetric.Consequently, the statement isTRUE.00610.0 pointsMake an orthogonal change of variablesthat reducesxTAx= 4x21+ 8x1x22x22= 0to a quadratic equation iny1, y2with no cross-product term given thatλ1λ2.1.6y21+ 4y22= 02.6y214y22= 0correct3.6y21+ 4y22=44.6y214y22=4Explanation:In matrix terms,xTAx= 4x21+ 8x1x22x22= [x1x2]4442x1x2.The eigenvaluesλ1,λ2ofAare the solutionsofdet4λ442λ=λ22λ24= (λ6)(λ+ 4) = 0,
Version 047 – NewEXAM04 – gilbert – (53525)4i.e.,λ1= 6 andλ2=4. Associated eigen-vectors areu1=21,u2=12,and these are orthogonal sinceλ1̸=λ2. Thenormalized eigenvectorsv1=1521,v2=1512,are thus orthonormal, andP= [v1v2] =152112is an orthogonal matrix such thatA=P6004P1=PDPTis an orthogonal diagonalization ofA.Now setx=x1x2,y=y1y2,x=Py.ThenxTAx= (Py)T(PDPT)Py=yT(PTP)D(PTP)y=yTDy=yT6004y= 6y214y22= 0.Consequently, whenx=Py,4x21+ 8x1x22x22= 6y214y22= 0.00710.0 pointsIfAT=Aand if vectorsuandvsatisfyAu= 3uandAv= 4v, thenu·v= 0.True or False?1.TRUEcorrect2.FALSEExplanation:The given vectorsuandvare eigenvectorsofAcorresponding to eigenvaluesλ1= 3 andλ2= 4. But whenAis symmetric, then anytwo eigenvectors from different eigenspacesare orthogonal.Thusuandvmust be or-thogonal, that is,u·v= 0.Consequently, the statement isTRUE.00810.0 pointsUse an orthogonal matrixPto identify2x24xy+ 5y2= 0
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