232
Linear Algebra, by Hefferon
(c) Iterates of this map cycle around
t
t
t
2
2
2
a + bx + cx2 b + cx + ax2 c + ax + bx2 a + bx + cx2
and the chains of range spaces and null spaces are trivial.
P2 =
246
Linear Algebra, by Hefferon
Five.IV.2.19 (a) The characteristic polynomial is c(x) = (x 3)2 and the minimal polynomial is the same.
(b) The characteristic polynomial is c(x) = (x + 1)2 . The minim
Answers to Exercises
247
(b) We want this matrix and its inverse.
P1
1
= 0
0
0
1
2
(c) The concatenation of these bases for the generalized
space.
1
1
0 0
B1 = 0 , 1
B3 =
1 0
0
1
3
4
0
null
248
Linear Algebra, by Hefferon
We took the rst basis vector so that it is in the null space of t2 but is not in the null space of t. The
second basis vector is the image of the rst under t.
(b) The c
Answers to Exercises
249
The null space of (t + 2)2 is the same, and so this is the generalized null space N (t + 2). Thus the
action of the restriction of t + 2 to N (t + 2) on an associated string b
250
Linear Algebra, by Hefferon
is nilpotent. The null spaces
x
z w
z w
z w
N ( t 6) = cfw_
x, z, w C
z, w C N (t 6)2 ) = cfw_
z
z
w
w
and the nullities show that the action of t 6 on
Jordan for
Answers to Exercises
251
Therefore there is only one Jordan form that is possible.
1 0 0
0 2 0
0 1 2
000
0
0
0
2
Five.IV.2.26 There are two possible Jordan forms. The action of t + 1 on a string basis
252
Linear Algebra, by Hefferon
Five.IV.2.28 The transformation d/dx : P3 P3 is nilpotent. Its action on B = x3 , 3x2 , 6x, 6 is x3
3x2 6x 6 0. Its Jordan form is its canonical form as nilpotent matr
10
Linear Algebra, by Hefferon
(b) The second components show that i = 2, the third components show that j = 1.
(c) m = 4, n = 2
One.I.2.21 For each problem we get a system of linear equations by look
Answers to Exercises
245
shows that f(A) is similar to f(B).
(a) Taking f to be a linear polynomial we have that A xI is similar to B xI. Similar matrices have
equal determinants (since |A| = |PBP1 |
244
Linear Algebra, by Hefferon
because the rst and second rows of the rst matrix D 3I are zero, the entire product will have a rst
row and second row that are zero. And because the third row of the m
Answers to Exercises
233
Five.III.2.19 (a) The domain has dimension four. The maps action is that any vector in the space
c1 1 + c2 2 + c3 3 + c4 4 goes to c1 2 + c2 0 + c3 4 + c4 0 = c1 3 + c3 4 . Th
234
Linear Algebra, by Hefferon
Np
p
1
2
2
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
1
1
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
0
0
0
1
0
0
0
0
N (Np )
r
u
cfw_ u v r, u, v C
u
v
r
s
cfw_ u v r, s, u,
Answers to Exercises
235
and P is the inverse of that.
P = RepE5 ,B (id) = (P1 )1
10
1 1
= 1 0
1 0
00
(c) The calculation to check this is routine.
Five.III.2.23
(a) The calculation
p
Np
1
0
0
1 1
1
0
236
Linear Algebra, by Hefferon
Five.III.2.24 A couple of examples
0
1
0
0
a
c
b
d
=
suggest that left multiplication by a
Distinct blocks
000
1 0 0
0 0 0
001
0
a
0
b
0
1
0
0
0
1
0
ab
0 d e
0
gh
0
c
f
Answers to Exercises
237
Five.III.2.30 We must check that B C cfw_ v1 , . . . , vj is linearly independent where B is a t-string basis for
R (t), where C is a basis for N (t), and where t(v1 ) = 1 ,
238
Linear Algebra, by Hefferon
Five.III.2.38 Some experimentation gives the idea for the proof. Expansion of the second power
t2 (T ) = S(ST T S) (ST T S)S = S2 2ST S + T S2
S
the third power
t3 (T )
Answers to Exercises
239
(a) Because T is triangular, T xI is also triangular
3x
0
0
T xI = 1
3x
0
0
0
4x
the characteristic polynomial is easy c(x) = |T xI| = (3 x)2 (4 x) = 1 (x 3)2 (x 4). There are
240
Linear Algebra, by Hefferon
Therefore, the minimal polynomial is m(x) = m3 (x) = (x 3)3 .
(d) This case is also triangular, here upper triangular.
c(x) = |T xI| =
2x
0
0
0
1
6x
2 = (2 x)2 (6 x) =
242
Linear Algebra, by Hefferon
and so the representation RepB,B (t) is this upper
1
0
0
0
triangular matrix.
111
1 2 3
0 1 3
001
Because it is triangular, the fact that the characteristic polynomial
Answers to Exercises
243
is (a x)(d x) bc = x2 (a + d)x + (ad bc). Substitute
2
a
c
b
d
( a + d)
a
c
b
d
+ (ad bc)
=
1
0
0
1
a2 + bc ab + bd
ac + cd bc + d2
a2 + ad
ac + cd
ab + bd
ad + d2
+
ad bc
0
Answers to Exercises
7
One.I.1.36 Solving the system
(1/3)(a + b + c) + d = 29
(1/3)(b + c + d) + a = 23
(1/3)(c + d + a) + b = 21
(1/3)(d + a + b) + c = 17
we obtain a = 12, b = 9, c = 3, d = 21. Thu
152
Linear Algebra, by Hefferon
Three.V.2.19 For reexivity, to show that any matrix is matrix equivalent to itself, take P and Q to be
identity matrices. For symmetry, if H1 = PH2 Q then H2 = P1 H1 Q1
160
Linear Algebra, by Hefferon
and then normalize.
1/ 2
1/ 6
1/ 2 , 1/6
0
2/ 6
Three.VI.2.13 Reducing the linear system
xyz+w=0
x
+z
=0
1 +2
xy z+w=0
y + 2z w = 0
and parametrizing gives this des
Answers to Exercises
161
or we just could modify the Gram-Schmidt procedure to throw out any zero vectors. The second thing
to worry about if we drop the phrase linearly independent from the question