Orthogonal Diagonalization

 

Selected Homework 10 Answers,  Exercise 3.

`Exercise 3.  Eigenvalue, eigenvector analysis of `*A*`:`

A = matrix([[2, -1, -1], [-1, 2, -1], [-1, -1, 2]]), `  `*(A-I*lambda) = matrix([[2-lambda, -1, -1], [-1, 2-lambda, -1], [-1, -1, 2-lambda]])

det(A-I*lambda)*` = `*lambda*(lambda-3)^2 = 0

evecsAval1 = 3

evecsAmul1 = 2

evecsAvec1 = vector([-1, 0, 1])

evecsAvec12 = vector([-1, 1, 0])

lambda = 3

A-I*lambda = matrix([[-1, -1, -1], [-1, -1, -1], [-1, -1, -1]])

matrix([[-1, -1, -1], [-1, -1, -1], [-1, -1, -1]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[1, 1, 1], [0, 0, 0], [0, 0, 0]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[a+b+c], [0], [0]]) = matrix([[0], [0], [0]])

v[1] = matrix([[-1], [0], [1]]), v[2] = matrix([[-1], [1], [0]])

evecsAval2 = 0

evecsAmul2 = 1

evecsAvec2 = vector([1, 1, 1])

lambda = 0

A-I*lambda = matrix([[2, -1, -1], [-1, 2, -1], [-1, -1, 2]])

matrix([[2, -1, -1], [-1, 2, -1], [-1, -1, 2]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[1, 0, -1], [0, 1, -1], [0, 0, 0]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[a-c], [b-c], [0]]) = matrix([[0], [0], [0]])

v[3] = matrix([[1], [1], [1]])

`A is diagonalizable using the eigenvectors of `*A*` to construct P:`

v[1] = matrix([[-1], [0], [1]]), v[2] = matrix([[-1], [1], [0]]), v[3] = matrix([[1], [1], [1]])

P = matrix([[-1, -1, 1], [0, 1, 1], [1, 0, 1]]), P^`-1` = matrix([[-1/3, -1/3, 2/3], [-1/3, 2/3, -1/3], [1/3, 1/3, 1/3]])

P^`-1`*P*` = `*matrix([[-1/3, -1/3, 2/3], [-1/3, 2/3, -1/3], [1/3, 1/3, 1/3]])*matrix([[-1, -1, 1], [0, 1, 1], [1, 0, 1]]) = matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

P^`-1`*A*P*` = `*matrix([[-1/3, -1/3, 2/3], [-1/3, 2/3, -1/3], [1/3, 1/3, 1/3]])*matrix([[2, -1, -1], [-1, 2, -1], [-1, -1, 2]])*matrix([[-1, -1, 1], [0, 1, 1], [1, 0, 1]]) = matrix([[3, 0, 0], [0, 3, ...

`Now find an orthogonal matrix Q that orthogonally diagonalizes A.`

`Begin with the three eigenvectors of A and convert to an orthonormal set: `

v[1] = vector([-1, 0, 1])

v[2] = vector([-1, 1, 0])

v[3] = vector([1, 1, 1])

`Apply Gram-Schmidt Process to these three vectors: `

w[1] = v[1]

w[2] = v[2]-`(`*v[2]*` dot`*` w`[1]*`)`/` ||`/w[1]/`||`^2*`w `[1]

w[3] = v[3]-`(`*v[3]*` dot`*` w`[1]*`)`/` ||`/w[1]/`||`^2*`w `[1]-`(`*v[3]*` dot`*` w`[2]*`)`/` ||`/w[2]/`||`^2*`w `[2]

w[1] = vector([-1, 0, 1])

w[2] = vector([-1, 1, 0])-`(`*vector([-1, 1, 0])*` dot`*vector([-1, 0, 1])*`)`/` ||`/vector([-1, 0, 1])/`||`^2*vector([-1, 0, 1])

w[3] = vector([1, 1, 1])-`(`*vector([1, 1, 1])*` dot`*vector([-1, 0, 1])*`)`/` ||`/vector([-1, 0, 1])/`||`^2*vector([-1, 0, 1])-`(`*vector([1, 1, 1])*` dot`*vector([-1/2, 1, -1/2])*`)`/` ||`/vector([-1...

`Obtain a basis of three orthogonal vectors: `

w[1] = vector([-1, 0, 1])

w[2] = vector([-1/2, 1, -1/2])

w[3] = vector([1, 1, 1])

`Obtain an orthonormal basis of three orthogonal unit vectors: `

u[1] = 1/` ||`/w[1]/`||`*` w`[1], `   ||`*w[1]*`||` = 2^(1/2)

u[2] = 1/` ||`/w[2]/`||`*` w`[2], `   ||`*w[2]*`||` = 1/2*6^(1/2)

u[3] = 1/` ||`/w[3]/`||`*` w`[3], `   ||`*w[3]*`||` = 3^(1/2)

u[1] = matrix([[-1/2*2^(1/2)], [0], [1/2*2^(1/2)]]), u[2] = matrix([[-1/6*6^(1/2)], [1/3*6^(1/2)], [-1/6*6^(1/2)]]), u[3] = matrix([[1/3*3^(1/2)], [1/3*3^(1/2)], [1/3*3^(1/2)]])

`Use these orthonormal column vectors to construct Q:`

Q = matrix([[-1/2*2^(1/2), -1/6*6^(1/2), 1/3*3^(1/2)], [0, 1/3*6^(1/2), 1/3*3^(1/2)], [1/2*2^(1/2), -1/6*6^(1/2), 1/3*3^(1/2)]]), Q^`-1` = matrix([[-1/2*2^(1/2), 0, 1/2*2^(1/2)], [-1/6*6^(1/2), 1/3*6^(...

Q^`-1`*Q*` = `*matrix([[-1/2*2^(1/2), 0, 1/2*2^(1/2)], [-1/6*6^(1/2), 1/3*6^(1/2), -1/6*6^(1/2)], [1/3*3^(1/2), 1/3*3^(1/2), 1/3*3^(1/2)]])*matrix([[-1/2*2^(1/2), -1/6*6^(1/2), 1/3*3^(1/2)], [0, 1/3*6^...

Q^`-1`*A*Q*` = `*matrix([[-1/2*2^(1/2), 0, 1/2*2^(1/2)], [-1/6*6^(1/2), 1/3*6^(1/2), -1/6*6^(1/2)], [1/3*3^(1/2), 1/3*3^(1/2), 1/3*3^(1/2)]])*matrix([[2, -1, -1], [-1, 2, -1], [-1, -1, 2]])*matrix([[-1...

`So, the symmetric matrix A is orthogonally diagonalizable using the orthogonal matrix Q.`