**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 s^{2} + 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 A^{9} 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 A**x**
= r**x**. Hence

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

r - s.

22.
The characteristic polynomial is r^{3} - 6r^{2} + 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^{-1}AP =

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 = s^{2}
- 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} = P^{T} so that P^{T}AP =

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) - 36^{2}]
= (s + 3)[s^{2 }+ 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} = P^{T} . 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 A^{T }is n by m so that A^{T}A is n
by n, i.e. is square.

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