Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 197
Subsection Five.IV.1: Polynomials of Maps and Matrices
Five.IV.1.13 For each, the minimal polynomial must have a leading coefficient of 1 and Theorem 1.8,
the Cayley-Hamilton Theorem, says that the minimal polynomial must contain the same linear factors
as the characteristic polynomial, although possibly of lower degree but not of zero degree.
(a) The possibilities are m 1 (x) = x − 3, m 2 (x) = (x − 3) 2 , m 3 (x) = (x − 3) 3 , and m 4 (x) = (x − 3) 4 .
Note that the 8 has been dropped because a minimal polynomial must have a leading coefficient of
one. The first is a degree one polynomial, the second is degree two, the third is degree three, and
the fourth is degree four.
(b) The possibilities are m 1 (x) = (x+1)(x−4), m 2 (x) = (x+1) 2 (x−4), and m 3 (x) = (x+1) 3 (x−4).
The first is a quadratic polynomial, that is, it has degree two. The second has degree three, and the
third has degree four.
(c) We have m 1 (x) = (x − 2)(x − 5), m 2 (x) = (x − 2) 2 (x − 5), m 3 (x) = (x − 2)(x − 5) 2 , and
m 4 (x) = (x − 2) 2 (x − 5) 2 . They are polynomials of degree two, three, three, and four.
(d) The possiblities are m 1 (x) = (x + 3)(x − 1)(x − 2), m 2 (x) = (x + 3) 2 (x − 1)(x − 2), m 3 (x) =
(x + 3)(x − 1)(x − 2) 2 , and m 4 (x) = (x + 3) 2 (x − 1)(x − 2) 2 . The degree of m 1 is three, the degree
of m 2 is four, the degree of m 3 is four, and the degree of m 4 is five.
Five.IV.1.14 In each case we will use the method of Example 1.12.
x
(a) Because T is triangular, T − xI is also triangular
⎛
T − xI = ⎝ 3 − x 1 3 0 − 0
0
⎞
⎠
0 0 4 − x
the characteristic polynomial is easy c(x) = |T −xI| = (3−x) 2 (4−x) = −1·(x−3) 2 (x−4). There are
only two possibilities for the minimal polynomial, m 1 (x) = (x−3)(x−4) and m 2 (x) = (x−3) 2 (x−4).
(Note that the characteristic polynomial has a negative sign but the minimal polynomial does not
since it must have a leading coefficient of one). Because m 1 (T ) is not the zero matrix
⎛
(T − 3I)(T − 4I) = ⎝ 0 0 0
⎞ ⎛
1 0 0⎠
⎝ −1 0 0
⎞ ⎛
1 −1 0⎠ = ⎝ 0 0 0
⎞
−1 0 0⎠
0 0 1 0 0 0 0 0 0
the minimal polynomial is m(x) = m 2 (x).
⎛
(T − 3I) 2 (T − 4I) = (T − 3I) · ((T
− 3I)(T − 4I) ) = ⎝ 0 0 0
⎞ ⎛
1 0 0⎠
⎝ 0 0 0
⎞ ⎛
−1 0 0⎠ = ⎝ 0 0 0
⎞
0 0 0⎠
0 0 1 0 0 0 0 0 0
(b) As in the prior item, the fact that the matrix is triangular makes computation of the characteristic
polynomial easy.
3 − x 0 0
c(x) = |T − xI| =
1 3 − x 0
∣ 0 0 3 − x∣ = (3 − x)3 = −1 · (x − 3) 3
There are three possibilities for the minimal polynomial m 1 (x) = (x − 3), m 2 (x) = (x − 3) 2 , and
m 3 (x) = (x − 3) 3 . We settle the question by computing m 1 (T )
⎛
T − 3I = ⎝ 0 0 0
⎞
1 0 0⎠
0 0 0
and m 2 (T ).
⎛
(T − 3I) 2 = ⎝ 0 0 0
⎞ ⎛
1 0 0⎠
⎝ 0 0 0
⎞ ⎛
1 0 0⎠ = ⎝ 0 0 0
⎞
0 0 0⎠
0 0 0 0 0 0 0 0 0
Because m 2 (T ) is the zero matrix, m 2 (x) is the minimal polynomial.
(c) Again, the matrix is triangular.
3 − x 0 0
c(x) = |T − xI| =
1 3 − x 0
∣ 0 1 3 − x∣ = (3 − x)3 = −1 · (x − 3) 3