Name ____Answers_____________                  Linear Algebra, Quiz 4, Summer 2004  

Problem 1. Find a matrix P that orthogonally diagonalizes A, use P to diagonalize A, and show that P is an orthogonal matrix.

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

det(A-I*lambda)*` = `*lambda^2*(lambda-1)*(lambda+1) = 0

evecsAval1 = 0

evecsAmul1 = 2

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

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

lambda = 0

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

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

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

matrix([[b], [d], [0], [0]]) = matrix([[0], [0], [0], [0]])

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

evecsAval2 = -1

evecsAmul2 = 1

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

lambda = -1

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

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

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

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

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

evecsAval3 = 1

evecsAmul3 = 1

evecsAvec3 = vector([0, 1, 0, 1])

lambda = 1

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

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

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

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

v[4] = matrix([[0], [1], [0], [1]])v[1] = matrix([[1], [0], [0], [0]]), v[2] = matrix([[0], [0], [1], [0]]), v[3] = matrix([[0], [-1], [0], [1]]), v[4] = matrix([[0], [1], [0], [1]])

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

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

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

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

v[4] = vector([0, 1, 0, 1])

Already orthogonal so don’t need Gram-Schmidt.

 w[1] = vector([1, 0, 0, 0])

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

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

w[4] = vector([0, 1, 0, 1])

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

u[1] = 1/` ||`/w[1]/`||`*` w`[1], `   ||`*w[1]*`||` = 1

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

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

u[4] = 1/` ||`/w[4]/`||`*` w`[4], `   ||`*w[4]*`||` = 2^(1/2)

u[1] = matrix([[1], [0], [0], [0]]), u[2] = matrix([[0], [0], [1], [0]]), u[3] = matrix([[0], [-1/2*2^(1/2)], [0], [1/2*2^(1/2)]]), u[4] = matrix([[0], [1/2*2^(1/2)], [0], [1/2*2^(1/2)]])

`Use these orthonormal column vectors to construct Q:`

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

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

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

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

Problem 2. Find a matrix P that orthogonally diagonalizes A, use P to diagonalize A, and show that P is an orthogonal matrix.

A = matrix([[-2/7, 6/7, 6/7], [6/7, 5/7, 0], [6/7, 0, 11/7]]), `  `*(A-I*lambda) = matrix([[-2/7-lambda, 6/7, 6/7], [6/7, 5/7-lambda, 0], [6/7, 0, 11/7-lambda]])

Expanding by the last row:

det(A-lI) = -686 + 343l + 686l2 - 343l3 = 0

det(A-I*lambda)*` = `*(lambda-1)*(lambda-2)*(lambda+1) = 0

evecsAval1 = 2

evecsAmul1 = 1

evecsAvec1 = vector([3/2, 1, 3])

lambda = 2

A-I*lambda = matrix([[-16/7, 6/7, 6/7], [6/7, -9/7, 0], [6/7, 0, -3/7]])

matrix([[-16/7, 6/7, 6/7], [6/7, -9/7, 0], [6/7, 0, -3/7]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[1, 0, -1/2], [0, 1, -1/3], [0, 0, 0]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[a-1/2*c], [b-1/3*c], [0]]) = matrix([[0], [0], [0]])

v[1] = matrix([[3/2], [1], [3]])

evecsAval2 = -1

evecsAmul2 = 1

evecsAvec2 = vector([-3, 3/2, 1])

lambda = -1

A-I*lambda = matrix([[5/7, 6/7, 6/7], [6/7, 12/7, 0], [6/7, 0, 18/7]])

matrix([[5/7, 6/7, 6/7], [6/7, 12/7, 0], [6/7, 0, 18/7]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[1, 0, 3], [0, 1, -3/2], [0, 0, 0]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[a+3*c], [b-3/2*c], [0]]) = matrix([[0], [0], [0]])

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

evecsAval3 = 1

evecsAmul3 = 1

evecsAvec3 = vector([1, 3, -3/2])

lambda = 1

A-I*lambda = matrix([[-9/7, 6/7, 6/7], [6/7, -2/7, 0], [6/7, 0, 4/7]])

matrix([[-9/7, 6/7, 6/7], [6/7, -2/7, 0], [6/7, 0, 4/7]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[1, 0, 2/3], [0, 1, 2], [0, 0, 0]])*matrix([[a], [b], [c]]) = matrix([[0], [0], [0]])

matrix([[a+2/3*c], [b+2*c], [0]]) = matrix([[0], [0], [0]])

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

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

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

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

v[1] = vector([3/2, 1, 3])

v[2] = vector([-3, 3/2, 1])

v[3] = vector([1, 3, -3/2])

Already orthogonal so don’t need Gram-Schmidt.

w[1] = vector([3/2, 1, 3])

w[2] = vector([-3, 3/2, 1])

w[3] = vector([1, 3, -3/2])

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

u[1] = 1/` ||`/w[1]/`||`*` w`[1], `   ||`*w[1]*`||` = 7/2

u[2] = 1/` ||`/w[2]/`||`*` w`[2], `   ||`*w[2]*`||` = 7/2

u[3] = 1/` ||`/w[3]/`||`*` w`[3], `   ||`*w[3]*`||` = 7/2

u[1] = matrix([[3/7], [2/7], [6/7]]), u[2] = matrix([[-6/7], [3/7], [2/7]]), u[3] = matrix([[2/7], [6/7], [-3/7]])

`Use these orthonormal column vectors to construct Q:`

Q = matrix([[3/7, -6/7, 2/7], [2/7, 3/7, 6/7], [6/7, 2/7, -3/7]]), Q^`-1` = matrix([[3/7, 2/7, 6/7], [-6/7, 3/7, 2/7], [2/7, 6/7, -3/7]])

Q^`-1`*Q*` = `*matrix([[3/7, 2/7, 6/7], [-6/7, 3/7, 2/7], [2/7, 6/7, -3/7]])*matrix([[3/7, -6/7, 2/7], [2/7, 3/7, 6/7], [6/7, 2/7, -3/7]]) = matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

Q^`-1`*A*Q*` = `*matrix([[3/7, 2/7, 6/7], [-6/7, 3/7, 2/7], [2/7, 6/7, -3/7]])*matrix([[-2/7, 6/7, 6/7], [6/7, 5/7, 0], [6/7, 0, 11/7]])*matrix([[3/7, -6/7, 2/7], [2/7, 3/7, 6/7], [6/7, 2/7, -3/7]]) = ...

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