Solution Key to Some Exercises in Chapter 3 #5. Determine the kernel and range of each of the following linear transformations on 2P
(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 2P into 2R defined by
10
()(())(0)p x dx p x p σ⎛⎫
⎪= ⎪⎝⎭
⎰ Find a matrix A such that
()x A ασαββ⎛⎫
+= ⎪⎝⎭
.
Solution
1(1)1σ⎛⎫
= ⎪⎝⎭ 1/2()0x σ⎛⎫
= ⎪⎝⎭
11/211/2()1010x ασαβαββ⎛⎫⎛⎫
⎛⎫⎛⎫
+=+= ⎪ ⎪
⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭
Hence, 11/210A ⎛⎫
=
⎪⎝⎭
#10. Let σ be the transformation on 3P defined by
(())'()"()p x xp x p x σ=+
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 1B S AS -=
d) If 2012()(1)p x a a x a x =+++, calculate (())n p x σ. Solution (a) (1)0σ=
()x x σ=
22()22x x σ=+
002010002A ⎛⎫
⎪
= ⎪ ⎪⎝⎭
(b) (1)0σ=
()x x σ=
22(1)2(1)x x σ+=+
000010002B ⎛⎫
⎪
= ⎪ ⎪⎝⎭
(c)
2[1,,1]x x +2[1,,]x x =101010001⎛⎫
⎪
⎪ ⎪⎝⎭
The transition matrix from 2[1,,]x x to 2[1,,1]x x + is
101010001S ⎛⎫ ⎪= ⎪ ⎪⎝⎭
, 1
B 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 1A P BP -=. Let 1S 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-L are linearly independent.
(b) there exists a basis E for V such that the matrix representing σ with respect to the basis E is
00001
0000
010⎛⎫
⎪
⎪
⎪ ⎪⎝⎭
L L M M M M L
Proof
(a) Suppose that
1011()()v v v 0n n k k k σσ--+++=L
Then 11011(()())v v v 0n n n k k k σσσ---+++=L
That is, 12210110()()())()v v v v 0n n n n n k k k k σσσσ----+++==L Thus, 0k must be zero since 1()v 0n σ-≠. 211111(()())()v v v 0n n n n k k k σσσσ----++==L
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 -L must be all zero.
This proves that
1,(),,()v v v n σσ-L are linearly independent.
(b) Since 1,(),,()v v v n σσ-L are n linearly independent, they form a basis for V .
Denote 112,(),,()εv εv εv n n σσ-===L 12()εεσ= 23()εεσ= …….
1()εεn n σ-= ()ε0n σ=
12[(),(),,()]εεεn σσσL 121[,,,,]εεεεn n -=L 00001
0000
010⎛⎫
⎪
⎪
⎪ ⎪⎝⎭
L L M M M M L
#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 λλλL . Then for each i , there exists a nonzero vector x i such that x x i i i A λ= x x x 0k k i i i i i A λλ=== since k A O =.
This will imply that 0i λ= for 1,2,,i n =L . 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 112diag(,,,)n A P P λλλ-=L then 112diag(,,,)k k k k n A P P λλλ-=L .
k A O = implies that 0i λ= for 1,2,,i n =L . Hence, B O =. This will imply that
A O =. Contradiction!。