標題:
orthogonal matrix
發問:
2. Let A =0 .....0..... ?2 ...............0.... ?2......0 ............?2...... 0...... 3 . (a) Find an orthogonal matrix P such that P power(?1) x A x P is diagonal. (b) Use (a) to show that A power(2n) is positive definite for all positive integers n.
最佳解答:
a) P must be formed by eigenvectors if A can be in form of P^(-1*)D*P, where D is a digonal matrix. To find an orthogonal matrix, you have to find the orthonormal eigenvector basis, i.e., eigenvectors with length 1 1) Find eigenvalue of A: (x=lamda, I=identity matrix) det(A-xI)=0 .......[-x........0.........-2] det...[0......-2-x........0].......=0 .......[-2.......0.........3-x] You can find that x=-1,-2,4 (due to text limit, the computation step is omitted) 2) Find (orthonormal) eigenvectors x=-1: [1.......0......-2|0].....................................[1......0.....-2|0] [0......-2.......0|0]......(after row reduce)......[0.....-2......0|0] [-2......0.......2|0].....................................[0.....0.......0|0] X2=0, X3=t, X1=2t....,hence (X1,X2,X3)=t(2,0,1)eigenvector=[2...0....1](t) Orthonormal eigenvector=[2/sqroot(5).....0......1/sqroot(5)](t) x=-2: [2.......0......-2|0].....................................[1......0.....-1|0] [0......-2.......0|0]......(after row reduce)......[0.....0.......3|0] [-2......0.......5|0].....................................[0.....0.......0|0] X2=0, X3=t, X1=2t....,hence (X1,X2,X3)=t(0,1,0)eigenvector=[0...1....0](t) Orthonormal eigenvector=[0...1....0](t) x=4: [-4.......0......-2|0].....................................[2......0......1|0] [0......-6........0|0]......(after row reduce)......[0.....1.......0|0] [-2......0.......-1|0].....................................[0.....0.......0|0] X2=0, X1=t, X3=-2t....,hence (X1,X2,X3)=t(1,0,-2)eigenvector=[1...0....-2](t) Orthonormal eigenvector=[1/sqroot(5).....0.......-2/sqroot(5)](t) Hence, P= [2/sqroot(5).....0.........1/sqroot(5)] [0..................1.................0......] [1/sqroot(5)....0.........-2/sqroot(5)] (If A is orthogonal matrix, then A*A(t)=I(identity matrix)...you can check that this is an orthogonal matrix) b) Apply theorem: If any matrix A can be in form of A=M^(-1)*D*M (where D is digonal matrix) and all of digonal entries are positive, then A must be positive definite from a), since A can be in form of P^(-1)DP, A^(n)=P^(-1)*D^(n)*P, and A^(2n)=P^(-1)*D^(2n)*P ..............[d11........0.........0] where D=[0...........d22......0] ..............[0..........0.......d33] .......................[...(d11^(2n)..........0...........0..........] and D^(2n) =.....[.......0..........d22^(2n)........0..........] .......................[.......0.................0............d33^(2n)] It is abvious that all entries of D must be positive for all positive integer n Hence, A^(2n) must be positive definite for all positive integer n
其他解答:
orthogonal matrix
發問:
2. Let A =0 .....0..... ?2 ...............0.... ?2......0 ............?2...... 0...... 3 . (a) Find an orthogonal matrix P such that P power(?1) x A x P is diagonal. (b) Use (a) to show that A power(2n) is positive definite for all positive integers n.
最佳解答:
a) P must be formed by eigenvectors if A can be in form of P^(-1*)D*P, where D is a digonal matrix. To find an orthogonal matrix, you have to find the orthonormal eigenvector basis, i.e., eigenvectors with length 1 1) Find eigenvalue of A: (x=lamda, I=identity matrix) det(A-xI)=0 .......[-x........0.........-2] det...[0......-2-x........0].......=0 .......[-2.......0.........3-x] You can find that x=-1,-2,4 (due to text limit, the computation step is omitted) 2) Find (orthonormal) eigenvectors x=-1: [1.......0......-2|0].....................................[1......0.....-2|0] [0......-2.......0|0]......(after row reduce)......[0.....-2......0|0] [-2......0.......2|0].....................................[0.....0.......0|0] X2=0, X3=t, X1=2t....,hence (X1,X2,X3)=t(2,0,1)eigenvector=[2...0....1](t) Orthonormal eigenvector=[2/sqroot(5).....0......1/sqroot(5)](t) x=-2: [2.......0......-2|0].....................................[1......0.....-1|0] [0......-2.......0|0]......(after row reduce)......[0.....0.......3|0] [-2......0.......5|0].....................................[0.....0.......0|0] X2=0, X3=t, X1=2t....,hence (X1,X2,X3)=t(0,1,0)eigenvector=[0...1....0](t) Orthonormal eigenvector=[0...1....0](t) x=4: [-4.......0......-2|0].....................................[2......0......1|0] [0......-6........0|0]......(after row reduce)......[0.....1.......0|0] [-2......0.......-1|0].....................................[0.....0.......0|0] X2=0, X1=t, X3=-2t....,hence (X1,X2,X3)=t(1,0,-2)eigenvector=[1...0....-2](t) Orthonormal eigenvector=[1/sqroot(5).....0.......-2/sqroot(5)](t) Hence, P= [2/sqroot(5).....0.........1/sqroot(5)] [0..................1.................0......] [1/sqroot(5)....0.........-2/sqroot(5)] (If A is orthogonal matrix, then A*A(t)=I(identity matrix)...you can check that this is an orthogonal matrix) b) Apply theorem: If any matrix A can be in form of A=M^(-1)*D*M (where D is digonal matrix) and all of digonal entries are positive, then A must be positive definite from a), since A can be in form of P^(-1)DP, A^(n)=P^(-1)*D^(n)*P, and A^(2n)=P^(-1)*D^(2n)*P ..............[d11........0.........0] where D=[0...........d22......0] ..............[0..........0.......d33] .......................[...(d11^(2n)..........0...........0..........] and D^(2n) =.....[.......0..........d22^(2n)........0..........] .......................[.......0.................0............d33^(2n)] It is abvious that all entries of D must be positive for all positive integer n Hence, A^(2n) must be positive definite for all positive integer n
其他解答:
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