Orthogonal Diagonalization

 

Selected Homework 10 Answers,  Exercise 2.

`Exercise 2.  Eigenvalue, eigenvector analysis of `*A*`:`

A = matrix([[1, 1, 0], [1, 1, 0], [0, 0, 0]]), `  `*(A-I*lambda) = matrix([[1-lambda, 1, 0], [1, 1-lambda, 0], [0, 0, -lambda]])

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

evecsAval1 = 0

evecsAmul1 = 2

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

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

lambda = 0

A-I*lambda = matrix([[1, 1, 0], [1, 1, 0], [0, 0, 0]])

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

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

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

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

evecsAval2 = 2

evecsAmul2 = 1

evecsAvec2 = vector([1, 1, 0])

lambda = 2

A-I*lambda = matrix([[-1, 1, 0], [1, -1, 0], [0, 0, -2]])

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

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

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

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

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

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

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

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

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

`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, 1, 0])

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

v[3] = vector([1, 1, 0])

`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, 1, 0])

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

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

`Obtain a basis of three orthogonal vectors: `

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

w[2] = vector([0, 0, 1])

w[3] = vector([1, 1, 0])

`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

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

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

`Use these orthonormal column vectors to construct Q:`

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

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

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

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