WeBWork Linear Algebra - Section 6.3 Diagonalization - Let A=-1,0,5,3.5-4 Find an invertible S and diagonal D such as that S^-1AS=D The characteristic

WeBWork Linear Algebra - Section 6.3 Diagonalization - Let...

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Let A=[-1,0,5,,3.5,-4]. Find an invertible S and diagonal D such as that S^-1AS=D. The characteristic equation of A is det(alphaI-A)=0 Consider: det(alphaI-A)=0 det(alpha+1,-0.5,,-3.5,alpha+4]=0 (alpha + 1)(alpha + 4)-(0.5)(-3.5)=0 alpha^2+5alpha+2.25=0 (alpha+0.5)(alpha+4.5)=0 alpha=-0.5,-4.5 Thus the eigenvalues of A are alpha1=-0.5,alpha2=-4.5 By definition, X=[x1,x2] is an eigenvector of A corresponding to alpha if and only X is a nontrivial solution of (alphaI-A)X=0. [alpha+1,-0.5,,-3.5,alpha+4][x1,x2]=[0,0] If alpha=-0.5 then [0.5, -0.5,,-3.5,3,5][x1,x2]=[0,0] [1,-1,,0,0][x1,x2]=[0,0] equation x1-x2=0 x2=t for some scalar t implies x1=t Thus, the eigenvector of A, corresponding to alpha=-0.5 are the nonzero vectors of the form X=[1,1]t, thus the eigenvector corresponding to alpha=-0.5 is [1,1] If alpha=-4.5 then [-3.5, -0.5,,-3.5,-0.5][x1,x2]=[0,0] [7,1,,0,0][x1,x2]=[0,0] equation 7x1+x2=0 Let x1=t for some scalar t, implies x2=-7t. Thus the eigenvector of A, corresponding to alpha=-4.5 are the nonzero vectors of the form [1,-7]t Thus the eigenvector corresponding to alpha=-4.5 is [1,-7] Let u1=[1,1] and u2=[1,-7] Take S=[1,1,,1,-7] S^-1 =[7/8,1/8,,1/8,-1/8] S^-1AS=[7/8,1/8,,1/8,-1/8][-1,0.5,,3.5,-4][1,1,,1,-7]=[-0.5,0,,0,-4.5]=D Therefore the matrices that satisfies S^-1AS=D are S=[1,1,,1,-7] and D=[-0.5,0,,0,-4.5] The matrix C=[-5,4,10,,8,-9,-20,,-4,4,9] has two distinct eigenvalues alpha1<alpha2: C=[-5,4,10,,8,-9,-20,,-4,4,9] Characteristic equation is det(A-alphaI)=[-5-alpha,4,10,,8,-9-alpha,-20,,-4,4,9-alpha]=(- 5-alpha)[(-9-alpha)(9-alpha)+80]-4(8(9-alpha))-80)+10(32+4(-9-alpha)) det(C-alphaI)=0 -alpha^3-5alpha^2-7alpha+3=0 (alpha+3)(alpha+1)^2 Here alpha1=-3 or alpha2=-1 Algebraic multiplicity of alpha1 is 1 Algebraic multiplicity of alpha2 is 2 Now find the eigenspace corresponding to the eigenvalue alpha1=-1
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det(C-(-1)I)=[-4,4,10,,8,-8,-20,-4,4,10]=[-4,4,10,,0,0,0,,0,0,0] Here it is a 3 by 3 matrix so n=3 linearly independent rows is r=1 Geometric multiplicity is n-r That is 3-1=2 Therefore, dimension of geometric space is 2. Here, another thing is AM=GM Algebraic multiplicity equals to geometric multiplicity Find a 2 x 2 matrix A for which E5=span[-3;5] and E-2=span[1;-2] where Ealpha is the eigenspace associated with the eigenvalue apha. A for which E(subscript -3) = span { [2, -5] } and E(subscript 6) = span { [-1, 3] }, let B = { i , j } be the basis of E and u = 2i - 5j v = -i + 3j , the eigenvectors, then Au = -3u and Av = 6v for A = (a c) (b d) this brings 4 equations with 4 unknows : 2a - 5c = -3*2 = -6 --------> eq (1) 2b - 5d =-3*(-5)= 15 --------> eq (2) -a + 3c = 6*(-1)= -6 --------> eq (3) -b + 3d = 6*3 = 18 --------> eq (4) 2* eq(3) gives -2a + 6c = -12 --------> eq(5) (1) + (5) gives : c = - 18 , if my calc is alright !
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