南航双语矩阵论matrixtheory第三章部分题解之欧阳学创编

Solution Key to Some Exercises in Chapter 3
#5.Determine the kernel and range of each of the following linear transformations on 2
P
(a) (())'()p x xp x σ= (b) (())()'()p x p x p x σ=- (c) (())(0)(1)p x p x p σ=+
Solution (a) Let ()p x ax b =+. (())p x ax σ=.
(())0p x σ= if and only if 0ax = if and only if 0a =. Thus,
ker(){|}b b R σ=∈
The range of σis 2
()P σ={|}ax a R ∈
(b) Let ()p x ax b =+. (())p x ax b a σ=+-.
(())0p x σ= if and only if 0ax b a +-= if and only if 0a =and
0b =.
Thus, ker(){0}σ=
The range of σis 2
()P σ=2
{|,}P ax b a a b R +-∈=
(c) Let ()p x ax b =+. (())p x bx a b σ=++.
(())0p x σ= if and only if 0bx a b ++= if and only if 0a =and
0b =.
Thus, ker(){0}σ=
The range of σis 2
()P σ=2
{|,}P
bx a b a b R ++∈=
备注: 映射的核以及映射的像都是集合,应该以集合的记号来表达或者用文字来叙述.
#7. Let be the linear mapping that maps 2
P into 2
R defined by
Find a matrix A such that ()x A ασαββ⎛⎫
+= ⎪
⎝⎭
.
Solution
Hence, 11/210A ⎛⎫= ⎪
⎝⎭
#10.Let σ be the transformation on 3
P defined by
a) Find the matrix A representing σ
with respect to 2[1,,]
x x
b) Find the matrix B representing σ with respect to
2[1,,1]
x x +
c) Find the matrix S such that 1
B S AS -= d) If 2012()(1)p x a a x a x =+++, calculate (())n
p x σ.
Solution (a) (1)0σ= (b) (1)0σ= (c)
The transition matrix from 2
[1,,]x x to 2
[1,,1]x x + is
101010001S ⎛⎫ ⎪= ⎪
⎪⎝⎭, 1B S AS -=
(d) 2201212((1))2(1)
n
n a a x a x a x a x σ
+++=++
#11. Let A and B be n n ⨯ matrices. Show that if A is similar to B then there exist n n ⨯ matrices S and T , with S nonsingular, such that
A ST =and
B TS =.
Proof There exists a nonsingular matrix P such that 1
A P BP -=.
Let 1
S P -=, T BP =. Then
A ST =and
B TS =.
#12.Let σ be a linear transformation on the vector space V of dimension n . If there exist a vector v such that 1
()v 0n σ-≠and ()v 0n
σ=, show that
(a)
1,(),,()
v v v n σσ- are linearly independent.
(b) there exists a basis E for V such that the matrix representing σwith respect to the basis E is Proof
(a)
Suppose that
Then 1
1011(()())v v v 0
n n n k k k σσσ---+++=
That is, 1
2210
110()()())()v v v v 0
n n n n n k k k k σ
σσσ----+++==
Thus, 0k must be zero since 1()v 0n σ-≠.
This will imply that 1k must be zero since 1
()v 0n σ-≠.
By repeating the process above, we obtain that
011,,
,n k k k - must be all zero. This proves that
1,(),
,()
v v v n σσ- are linearly independent.
(b) Since 1
,(),,()
v v v n σσ- are n linearly independent, they
form a basis for V . Denote 112,(),
,()
εv εv εv n n σσ-===
…….
#13. If A is a nonzero square matrix and
k A O
=for some
positive integer k , show that A can not be similar to a diagonal matrix.
Proof Suppose that A is similar to a diagonal matrix
12diag(,,
,)n λλλ. Then for each i , there exists a nonzero vector x i
such that x x i
i i A λ=
x x x 0
k k i i i i i A λλ=== since k
A
O =.
This will imply that 0
i
λ
= for 1,2,
,i n
=. Thus, matrix A is
similar to the zero matrix. Therefore, A O =since a matrix that is similar to the zero matrix must be the zero matrix, which contradicts the assumption.
This contradiction shows that A can not be similar to a diagonal matrix. Or If
1
12diag(,,
,)n A P P λλλ-= then
1
12diag(,,,)k k k k n A P P λλλ-=.
k A O = implies
that 0i
λ= for 1,2,,i n =. Hence, B O =. This will
imply that A O =. Contradiction!。