3.1 Linear Transformations

从这一章开始,就讨论本书的主题线性变换了。定义很像subspace的定义,线性变换是一个满足 T ( c α + β ) = c T ( α ) + T ( β ) T(c\alpha+\beta)=cT(\alpha)+T(\beta) 的函数。常见的例子有zero transformation,differentiation transformation,矩阵乘法的transformation,积分transformation等。
如果 T T 是一个linear transformation,那么 T ( 0 ) = 0 T(0)=0 ,且 T T 保持linear combination。Theorem 1说明 V V 的一组基可以用唯一的线性变换指到 W W 的任何一组 dim V \dim V 数量的向量上,特别地,任意一个linear transformation T T 是唯一地由其 images of standard basis 决定。并且可以使用由 images of standard basis 为行组成的矩阵来显式表示出来。
接下来讨论的是range和null space,二者都是subspace,由定义可以证。range的dimension称为rank of T,null space的dimension称为nullity of T。Theorem 2是线性代数中一个很重要的定理: r a n k ( T ) + n u l l i t y ( T ) = dim V rank(T)+nullity(T)=\dim V 。由此可以得到关于矩阵的Theorem 3:矩阵的行rank和列rank相同。证明中要使用到方程组解空间的维数是 n r n-r ,其中 r r 为系数矩阵行空间dimension的结论。

Exercises

1. Which of the following functions T T from R 2 R^2 into R 2 R^2 are linear transformations?
( a ) T ( x 1 , x 2 ) = ( 1 + x 1 , x 2 ) T(x_1,x_2)=(1+x_1,x_2)
( b ) T ( x 1 , x 2 ) = ( x 2 , x 1 ) T(x_1,x_2)=(x_2,x_1)
( c ) T ( x 1 , x 2 ) = ( x 1 2 , x 2 ) T(x_1,x_2)=(x_1^2,x_2)
( d ) T ( x 1 , x 2 ) = ( sin x 1 , x 2 ) T(x_1,x_2)=(\sin x_1,x_2)
( e ) T ( x 1 , x 2 ) = ( x 1 x 2 , 0 ) T(x_1,x_2)=(x_1-x_2,0)

Solution:
( a ) No, since T ( 2 ( 1 , 0 ) ) = T ( 2 , 0 ) = ( 1 + 2 , 0 ) = ( 3 , 0 ) 2 ( 2 , 0 ) = 2 T ( 1 , 0 ) T(2(1,0))=T(2,0)=(1+2,0)=(3,0)\neq 2(2,0)=2T(1,0) .
( b ) Yes, since
T ( c ( x 1 , x 2 ) + ( y 1 , y 2 ) ) = ( c x 2 + y 2 , c x 1 + y 1 ) = c ( x 2 , x 1 ) + ( y 2 , y 1 ) = c T ( x 1 , x 2 ) + T ( y 1 , y 2 ) \begin{aligned}T(c(x_1,x_2)+(y_1,y_2))&=(cx_2+y_2,cx_1+y_1)\\&=c(x_2,x_1)+(y_2,y_1)\\&=cT(x_1,x_2)+T(y_1,y_2)\end{aligned}
( c ) No.
( d ) No.
( e ) Yes, since
T ( c ( x 1 , x 2 ) + ( y 1 , y 2 ) ) = ( c x 1 + y 1 c x 2 y 2 , 0 ) = c ( x 1 x 2 , 0 ) + ( y 1 y 2 , 0 ) = c T ( x 1 , x 2 ) + T ( y 1 , y 2 ) \begin{aligned}T(c(x_1,x_2)+(y_1,y_2))&=(cx_1+y_1-cx_2-y_2,0)\\&=c(x_1-x_2,0)+(y_1-y_2,0)\\&=cT(x_1,x_2)+T(y_1,y_2)\end{aligned}

2. Find the range, rank, null space, and nullity for the zero transformation and the identity transformation on a finite-dimensional space V V .

Solution: Let Z Z be the linear transformation, and I I be the identity transformation, then
the range of Z Z is 0, the rank of Z Z is 0, the null space of Z Z is V V , the nullity of Z Z is dim V \dim V .
the range of I I is V V , the rank of I I is dim V \dim V , the null space of I I is 0, the nullity of I I is 0.

3. Describe the range and the null space for the differentiation transformation of Example 2. Do the same for the integration transformation of Example 5.

Solution:
For Example 2, range D = V D=V , null D = { f ( x ) : f ( x ) = k , k F } D=\{f(x):f(x)=k,k\in F\} .
For Example 5, null T = { 0 } T=\{0\} , range T = { f : f  has a continuous first derivative } T=\{f:f \text{ has a continuous first derivative}\} .

4. Is there a linear transformation T T from R 3 R^3 into R 2 R^2 such that T ( 1 , 1 , 1 ) = ( 1 , 0 ) T(1,-1,1)=(1,0) and T ( 1 , 1 , 1 ) = ( 0 , 1 ) T(1,1,1)=(0,1) ?

Solution: Let T T be defined as
T ( 1 , 1 , 1 ) = ( 1 , 0 ) , T ( 1 , 1 , 1 ) = ( 0 , 1 ) , T ( 0 , 0 , 1 ) = ( 0 , 0 ) T(1,-1,1)=(1,0),\quad T(1,1,1)=(0,1),\quad T(0,0,1)=(0,0)
Since ( 1 , 1 , 1 ) , ( 1 , 1 , 1 ) , ( 0 , 0 , 1 ) (1,-1,1),(1,1,1),(0,0,1) is a basis for R 3 R^3 , T T is well defined.

5. If

α 1 = ( 1 , 1 ) , β 1 = ( 1 , 0 ) α 2 = ( 2 , 1 ) , β 2 = ( 0 , 1 ) α 3 = ( 3 , 2 ) , β 3 = ( 1 , 1 ) \alpha_1=(1,-1),\quad \beta_1=(1,0) \\ \alpha_2=(2,-1),\quad \beta_2=(0,1) \\\alpha_3=(-3,2),\quad \beta_3=(1,1)

is there a linear transformation T T from R 2 R^2 into R 2 R^2 such that T α i = β i T\alpha_i=\beta_i for i = 1 , 2 , 3 i=1,2,3 ?

Solution: Assume such a T T exists, then T ( α 1 α 2 ) = T α 1 T α 2 T(-\alpha_1-\alpha_2 )=-T\alpha_1-T\alpha_2 , notice that α 1 α 2 = α 3 -\alpha_1-\alpha_2=\alpha_3 , we have
β 3 = T α 3 = T α 1 T α 2 = β 1 β 2 = ( 1 , 1 ) = β 3 \beta_3=T\alpha_3=-T\alpha_1-T\alpha_2=-\beta_1-\beta_2=-(1,1)=-\beta_3
this means β 3 = ( 0 , 0 ) \beta_3=(0,0) , a contradiction.

6. Describe explicitly (as in Exercises 1 and 2) the linear transformation T T from F 2 F^2 into F 2 F^2 such that T ϵ 1 = ( a , b ) , T ϵ 2 = ( c , d ) T\epsilon_1=(a,b),T\epsilon_2=(c,d) .

Solution:
T ( x 1 , x 2 ) = x 1 T ϵ 1 + x 2 T ϵ 2 = x 1 ( a , b ) + x 2 ( c , d ) = ( x 1 a + x 2 c , x 1 b + x 2 d ) \begin{aligned}T(x_1,x_2 )&=x_1 T\epsilon_1+x_2 T\epsilon_2=x_1 (a,b)+x_2 (c,d)\\&=(x_1 a+x_2 c,x_1 b+x_2 d)\end{aligned}
If a = b = c = d = 0 a=b=c=d=0 , then T T is the zero transformation.
If at least one of a , b , c , d a,b,c,d is not 0, but a d b c = 0 ad-bc=0 , then the rank of T T and the nullity of T T are both 1. The range of T T is { ( 0 , y ) : y R } \{(0,y):y\in R\} if a = c = 0 a=c=0 , { ( x , 0 ) : x R } \{(x,0):x\in R\} if b = d = 0 b=d=0 , { k ( c , d ) : k R } \{k(c,d):k\in R\} if a = b = 0 a=b=0 , { k ( a , b ) : k R } \{k(a,b):k\in R\} if c = d = 0 c=d=0 , and { k ( c , d ) : k R } \{k(c,d):k\in R\} otherwise.
If a d b c 0 ad-bc\neq 0 , then the rank of T T is 2 and the nullity of T T is 0, The range of T T is R 2 R^2 , the null space of T T is { 0 } \{0\} .

7. Let F F be a subfield of the complex numbers and let T T be the function from F 3 F^3 into F 3 F^3 defined by

T ( x 1 , x 2 , x 3 ) = ( x 1 x 2 + 2 x 3 , 2 x 1 + x 2 , x 1 2 x 2 + 2 x 3 ) . T(x_1,x_2,x_3)=(x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3).

( a ) Verify that T T is a linear transformation.
( b ) If ( a , b , c ) (a,b,c) is a vector in F 3 F^3 , what are the conditions on a , b , c a,b,c that the vectors be in the range of T T ? What is the rank of T T ?
( c ) What are the conditions on on a , b , c a,b,c that ( a , b , c ) (a,b,c) be in the null space of T T ? What is the nullity of T T ?

Solution:
( a ) Let x = ( x 1 , x 2 , x 3 ) , y = ( y 1 , y 2 , y 3 ) x=(x_1,x_2,x_3),y=(y_1,y_2,y_3) , then it’s easy to see
T ( c x + y ) = T ( c x 1 + y 1 , c x 2 + y 2 , c x 3 + y 3 ) = c T x + T y T(cx+y)=T(cx_1+y_1,cx_2+y_2,cx_3+y_3 )=cTx+Ty
( b ) The conditions are a b = c a-b=c , the rank of T T is 2.
( c ) ( a , b , c ) = ( 2 t , 4 t , 3 t ) , t R (a,b,c)=(2t,-4t,-3t),t\in R , the nullity of T T is 1.

8. Describe explicitly a linear transformation from R 3 R^3 into R 3 R^3 which has as its range the subspace spaned by ( 1 , 0 , 1 ) (1,0,-1) and ( 1 , 2 , 2 ) (1,2,2) .

Solution: One possible choice of T T can be
T ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 , 2 x 2 + 2 x 3 , x 1 + 2 x 2 + 3 x 3 ) T(x_1,x_2,x_3 )=(x_1+x_2,2x_2+2x_3,-x_1+2x_2+3x_3)
Since T ϵ 1 = ( 1 , 0 , 1 ) , T ϵ 2 = ( 1 , 2 , 2 ) , T ϵ 3 = ( 0 , 2 , 3 ) = T ϵ 2 T ϵ 1 T\epsilon_1=(1,0,-1),T\epsilon_2=(1,2,2),T\epsilon_3=(0,2,3)=T\epsilon_2-T\epsilon_1 , the condition is satisfied.

9. Let V V be the vector space of all n × n n\times n matrices over the field F F , and let B B be a fixed n × n n\times n matrix. If

T ( A ) = A B B A T(A)=AB-BA

verify that T T is a linear transformation from V V into V V .

Solution: Let A , D A,D be any matrix in V V , then
T ( c A + D ) = ( c A + D ) B B ( c A + D ) = c A B + D B c B A B D = c ( A B B A ) + ( D B B D ) = c T ( A ) T ( D ) \begin{aligned}T(cA+D)&=(cA+D)B-B(cA+D)\\&=cAB+DB-cBA-BD\\&=c(AB-BA)+(DB-BD)\\&=cT(A)-T(D)\end{aligned}

10. Let V V be the set of all complex numbers regarded as a vector space over the field of real numbers (usual operations). Find a function from V V into V V which is a linear transformation on the above vector space, but which is not a linear transformation on C 1 C^1 , i.e., which is not complex linear.

Solution: Let f be defined as f ( a + b i ) = a f(a+bi)=a for a + b i C a+bi\in C (in which a , b R a,b\in R ), then f f is a linear transformation from V V into V V . To see it’s not a transformation on C 1 C^1 , we have f ( 1 + i ) = 1 f(1+i)=1 , and f ( i ) = 0 f(i)=0 , but for c = 1 + i c=1+i we have
f ( c ( 1 + i ) + i ) = f ( 3 i ) = 0 c f ( 1 + i ) + f ( i ) f(c(1+i)+i)=f(3i)=0\neq cf(1+i)+f(i)

11. Let V V be the space of n × 1 n\times 1 matrices over F F and let W W be the space of m × 1 m\times 1 matrices over F F . Let A A be a fixed m × n m\times n matrix over F F and let T T be the linear transformation from V V into W W defined by T ( X ) = A X T(X)=AX . Prove that T T is the zero transformation if and only if A A is the zero matrix.

Solution: If A A is the zero matrix, then obviously T T is the zero transformation. Now suppose T T is the zero transformation and assume A 0 A\neq 0 , then at least one a i j 0 a_{ij}\neq 0 , here 1 i m , 1 j n 1\leq i\leq m,1\leq j\leq n , we let X = a i j ϵ j X=a_{ij} \epsilon_j in V V , the j j -th coordinate of X X is a i j a_{ij} and the other coordinates are 0 0 , then the i i -th coordinate of A X AX is a i j 2 0 a_{ij}^2\neq 0 , thus T ( X ) = A X 0 T(X)=AX\neq 0 , a contradiction.

12. Let V V be an n n -dimensional vector space over the field F F and let T T be a linear transformation from V V into V V such that the range and null space of T T are identical. Prove that n n is even. (Can you give an example of such a linear transformation T T )?

Solution: We have r a n k ( T ) = n u l l i t y ( T ) rank(T)=nullity(T) , since the range of T T and the null space of T T are equal, thus
n = dim V = r a n k ( T ) + n u l l i t y ( T ) = 2 r a n k ( T ) n=\dim V=rank(T)+nullity(T)=2rank(T)
so n n is even. One example may be T ( x 1 , x 2 ) = ( x 2 , 0 ) T(x_1,x_2 )=(x_2,0) .

13. Let V V be a vector space over the field F F and T T a linear transformation from V V into V V . Prove that the following two statements about T T are equivalent.
( a ) The intersection of the range of T T and the null space of T T is the zero subspace of V V .
( b ) If T ( T α ) = 0 T(T\alpha)=0 , then T ( α ) = 0 T(\alpha)=0 .

Solution: Let A A be the range of T T and B B be the null space of T T .
First suppose (a) is true, then A B = { 0 } A\cap B=\{0\} , if T ( T α ) = 0 T(T\alpha)=0 , then T α B T\alpha\in B , but obviously T α A T\alpha\in A , thus T α A B T\alpha\in A\cap B and T α = 0 T\alpha=0 .
Conversely, suppose T ( T α ) = 0 T(T\alpha)=0 means T α = 0 T\alpha=0 , and let x A B x\in A\cap B , then T x = 0 Tx=0 , and there’s β V \beta\in V s.t. T β = x T\beta=x , so we have T ( T β ) = T x = 0 T(T\beta)=Tx=0 , so T β = 0 = x T\beta=0=x , thus A B = { 0 } A\cap B=\{0\} .

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3.1