MATH1025 AF


FW00F


Answers to Tutorial 9



Anton and Rorres


Section 7.1


See text for answers to questions 1, 2 and 3. However, note that in part (d) the characteristic polynomial is s2 + 3 = 0 which has complex roots with corresponding eigenvectors Note that you are expected to solve for complex eigenvalues and eigenvectors.


11. Since A is a triangular matrix, its eigenvalues are just the diagonal elements. The eigenvalues of A9 are just the ninth power of the eigenvalues of A - see text for values.


14. The constant term in the characteristic polynomial is (-1)n det(A) where n is the size of the matrix. See text for answers.


21. If r is an eigenvalue of A with eigenvector x then Ax = rx. Hence

(A - sI)x = Ax - sx = rx - sx = (r - s)x, i.e. x is an eigenvector of A - sI with eigenvalue

r - s.


22. The characteristic polynomial is r3 - 6r2 + 11r - 6 which has roots 1, 2 and 3 with corresponding eigenvectors (1, 0, 1), (1, 2 , 0) and (1, 1, 1).

(a) A-1 has eigenvalues which are the inverse of the eigenvalues of A with the same eigenvectors;

(b) A - 3I has eigenvalues which are 3 less than those of A with the same eigenvectors;

(c) A + 2I has eigenvalues which are 2 more than those of A with the same eigenvectors.

See text for detailed answers.

Section 7.2


2. (a) The characteristic polynomial is (3 - s)[(4 - s)2 - 1] which has roots s = 3 with multiplicity 2 and s = 5.

(b) For s = 3, the matrix sI - A is which has reduced row echleon form

so that it has rank 1 and nullity 2 which means the eigenspace has two basis vectors.

For s = 5, the matrix sI - A is which has reduced row echleon form

so that it has rank 2 and nullity 1 which means the eigenspace has one basis vector.

(c) Since A is square with dimension 3 and has 3 linearly independent eigenvectors, the basis vectors of its eigenspaces, it is diagonalizable.


5. Since the matrix is triangular, its eigenvalues are 3 and 2 with multiplicty 2.

For s = 3, sI- A is which has rank 2 and nullity 1.

For s = 2, sI - A is which has rank 2 and nullity 1.

Thus the matrix has only two linearly independent eigenvalues and is not diagonalizable.


10. The matrix has characteristic polynomial (s - 1)[(s - 1)2 - 1] which has roots 0, 1 and 2. Since it has three distinct roots, it has three linearlyindependent eigenvectors and hence is diagonalizable.

For s = 0, A has an eigenvector (0, 1, -1).

For s = 1, A has an eigenvector (1, 0, 0).

For s = 2, A has an eigenvector (0, 1, 1)

Taking these eigenvectors as the columns of P we get P = so that

AP =. Now P-1 = so that P-1AP =


14. Since A is triangular, it has one eigenvalue 5 with multiplicity 3. Then 5I - A is

which has rank 2 and nullity 1. Thus it has only one linearly independent eigenvalue and hence it is not diagonalizable.

Section 7.3


4. The matrix has characteristic polynomial (s - 6)(s - 3) - 4 = s2 - 9s + 14 with roots 2 and 7.

The normalized eigenvector corresponding to s = 2 is

The normalized eigenvector corresponding to s = 7 is

Thus P = and P-1 = PT so that PTAP =

Note that the answer in the text has solved for the eigenvectors in the opposite order and has got a different answer but both are correct. That is, the diagonalization is not unique.


5. The characteristic polynomial is (s + 3)[(s + 2)(s + 23) - 362] = (s + 3)[s2 + 25s -1250] which has roots -3, -50 and 25.

The normalized eigenvector corresponding to s = 25 is

The normalized eigenvector corresponding to s = -3 is

The normalized eigenvector corresponding to s = -50 is

Thus P = and P-1 = PT . See answer in text. Again we have chosen a different form for P but got the same diagonal matrix.


11.If A is m by n then AT is n by m so that ATA is n by n, i.e. is square.

Also (ATA)T = ATA so ATA is symmetric and hence has an orthonormal set of eigenvectors by Thm 7.3.1.