2 ? ? 2 ? 6 the eigenvalues of a are the solutions of

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)(2 +λ) =λ2+λ-6,the eigenvalues ofAare the solutions ofλ2+λ-6 = (λ+ 3)(λ-2) = 0,i.e.,λ=-3,2.ThusAis diagonalizablebecause the eigenvalues ofAare distinct, andA=PDP-1withP= [v1v2]wherev1andv2are eigenvectors correspond-ing toλ1andλ2respectively.To determinev1andv2we solve the equationAx=λx.Ax=λx, λ=-3:bracketleftbigg141-2bracketrightbigg bracketleftbiggx1x2bracketrightbigg=bracketleftbiggx1+ 4x2x1-2x2bracketrightbigg=-3bracketleftbiggx1x2bracketrightbigg,which can be written asx1+ 4x2=-3x1,x1-2x2=-3x2,i.e.,x1=-x2. So one choice ofv1isv1=bracketleftbigg-11bracketrightbigg.Ax=λx, λ= 2:bracketleftbigg141-2bracketrightbigg bracketleftbiggx1x2bracketrightbigg=bracketleftbiggx1+ 4x2x1-2x2bracketrightbigg= 2bracketleftbiggx1x2bracketrightbigg,which can be written asx1+ 4x2= 2x1,x1-2x2= 2x2,i.e.,x1= 4x2. So one choice ofv2isv2=bracketleftbigg41bracketrightbigg.Consequently,A=PDP-1withP=bracketleftbigg-1411bracketrightbigg.00910.0pointsThe eigenvalues of the matrixA=20-2132003areλ= 2,3,3.IfAis diagonalizable,i.e.,A=PDP-1withPinvertible andDdiagonal, which ofthe following is a choice forP?1.Ais not diagonalizable2.P=10-21100013.P=-10-2110001correct4.P=-1021100015.P=102110001Explanation:We first determine the eigenspaces corre-sponding toλ= 2,3:λ= 2 : sincerref(A-2I) = rref00-2112001=110001000,there is only free variable andNul(A-2I) =s-110:sinR.
cruz (fmc326) – HW09 – gilbert – (56540)6Thus Nul(A-2I) has dimension 1, andv1=-110is a basis for Nul(A-2I).λ= 3 : sincerref(A-3I) = rref-10-2102000=102000000,there are two free variables andNul(A-3I) =s010+t-201:s, tinR.Thus Nul(A-3I) has dimension 2, andv2=010,v3=-201,form a basis for Nul(A-3I).Consequently,Ais diagonalizable becausev1,v2,v3are linearly independent, andA=PDP-1withP= [v1v2v2] =-10-2110001.01010.0pointsIf eigenvectors of ann×nmatrixAare abasis forRn, thenAis diagonalizable.True or False?1.FALSE2.TRUEcorrectAnn×nmatrixAis diagonalizable if andonly ifAhasnlinearly independent eigenvec-tors. On the other hand,nvectors inRnarea basis forRnif and only if the vectors arelinearly independent.So if eigenvectors ofAform a basis forRn,they must be linearly independent, in whichcaseAwill be diagonalizable.Consequently, the statement isTRUE.01110.0pointsIfAis a diagonalizablen×nmatrix, thenAis invertible.True or False?1.TRUE2.FALSEcorrectExplanation:Consider the 3×3 triangular matrixA=5-8100700-2BecauseAis triangular, its eigenvalues are theentries along the diagonal,i.e.,λ= 5,0,-2.Since these are distinct,Ais diagonalizable.On the other hand, one of its eigenvaluesis zero, soAis not invertible (or note thatdet[A] = 0 because det[A] is the product5(0)(-2) = 0 of the diagonal values ofA,hence not invertible).Therefore, ann×nExplanation:

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