南航双语矩阵论 matrix theory第三章部分题解

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()101

0x 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 ⎛⎫ ⎪= ⎪ ⎪⎝⎭

, 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 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- are linearly independent.

(b) there exists a basis E for V such that the matrix representing σ with respect to the basis E is

000010000010⎛⎫ ⎪ ⎪ ⎪ ⎪⎝⎭

Proof

(a) Suppose that

1011()()v v v 0n n k k k σσ--++

+= Then 11011(()())v v v 0n n n k k k σσσ---+++=

That is, 12210110()()())()v v v v 0n n n n n k k k k σσσσ----++

+== Thus, 0k must be zero since 1()v 0n σ-≠.

211111(()())()v v v 0n n n n k k k σσσσ----++==

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-===

12()εεσ=

23()εεσ=

…….

1()εεn n σ-=

()ε0n σ=

12[(),(),,()]εεεn σσσ121[,,,,]εεεεn n -=000010000010⎛⎫ ⎪ ⎪ ⎪ ⎪⎝⎭

#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 0k k i i i i i A λλ=== since k A O =.

This will imply that 0i 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