Orthogonal Diagonalization

Example 4.  (Eigenvalues & Eigenvectors Example 6.)

`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([[4, 2, 2], [2, 4, 2], [2, 2, 4]])*matrix([[-1, -1, 1], [0, 1, 1], [1, 0, 1]]) = matrix([[2, 0, 0], [0, 2, 0], [0...

``

`Is A also orthogonally diagonalizable ?`

`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([[4, 2, 2], [2, 4, 2], [2, 2, 4]])*matrix([[-1/2*2^(...

`Note that A is orthogonally diagonalizable using Q, which is what we expect since A is symmetric.`