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(cα+β)=cT(α)+T(β)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是一个linear transformation,那么T(0)=0,且T保持linear combination。Theorem 1说明V的一组基可以用唯一的线性变换指到W的任何一组dimV数量的向量上,特别地,任意一个linear transformation 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:rank(T)+nullity(T)=dimV。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,其中r为系数矩阵行空间dimension的结论。
Exercises
1. Which of the following functions T from R2 into R2 are linear transformations?
( a ) T(x1,x2)=(1+x1,x2)
( b ) T(x1,x2)=(x2,x1)
( c ) T(x1,x2)=(x12,x2)
( d ) T(x1,x2)=(sinx1,x2)
( e ) T(x1,x2)=(x1−x2,0)
Solution: ( a ) No, since T(2(1,0))=T(2,0)=(1+2,0)=(3,0)=2(2,0)=2T(1,0). ( b ) Yes, since T(c(x1,x2)+(y1,y2))=(cx2+y2,cx1+y1)=c(x2,x1)+(y2,y1)=cT(x1,x2)+T(y1,y2) ( c ) No. ( d ) No. ( e ) Yes, since T(c(x1,x2)+(y1,y2))=(cx1+y1−cx2−y2,0)=c(x1−x2,0)+(y1−y2,0)=cT(x1,x2)+T(y1,y2)
2. Find the range, rank, null space, and nullity for the zero transformation and the identity transformation on a finite-dimensional space V.
Solution: Let Z be the linear transformation, and I be the identity transformation, then the range of Z is 0, the rank of Z is 0, the null space of Z is V, the nullity of Z is dimV. the range of I is V, the rank of I is dimV, the null space of I is 0, the nullity of 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, null D={f(x):f(x)=k,k∈F}. For Example 5, null T={0}, range T={f:f has a continuous first derivative}.
4. Is there a linear transformation T from R3 into R2 such that T(1,−1,1)=(1,0) and T(1,1,1)=(0,1)?
Solution: Let T be defined as T(1,−1,1)=(1,0),T(1,1,1)=(0,1),T(0,0,1)=(0,0) Since (1,−1,1),(1,1,1),(0,0,1) is a basis for R3, T is well defined.
is there a linear transformation T from R2 into R2 such that Tαi=βi for i=1,2,3?
Solution: Assume such a T exists, then T(−α1−α2)=−Tα1−Tα2, notice that −α1−α2=α3, we have β3=Tα3=−Tα1−Tα2=−β1−β2=−(1,1)=−β3 this means β3=(0,0), a contradiction.
6. Describe explicitly (as in Exercises 1 and 2) the linear transformation T from F2 into F2 such that Tϵ1=(a,b),Tϵ2=(c,d).
Solution: T(x1,x2)=x1Tϵ1+x2Tϵ2=x1(a,b)+x2(c,d)=(x1a+x2c,x1b+x2d) If a=b=c=d=0, then T is the zero transformation. If at least one of a,b,c,d is not 0, but ad−bc=0, then the rank of T and the nullity of T are both 1. The range of T is {(0,y):y∈R} if a=c=0, {(x,0):x∈R} if b=d=0, {k(c,d):k∈R} if a=b=0, {k(a,b):k∈R} if c=d=0, and {k(c,d):k∈R} otherwise. If ad−bc=0, then the rank of T is 2 and the nullity of T is 0, The range of T is R2, the null space of T is {0}.
7. Let F be a subfield of the complex numbers and let T be the function from F3 into F3 defined by
( b ) If (a,b,c) is a vector in F3, what are the conditions on a,b,c that the vectors be in the range of T? What is the rank of T?
( c ) What are the conditions on on a,b,c that (a,b,c) be in the null space of T? What is the nullity of T?
Solution: ( a ) Let x=(x1,x2,x3),y=(y1,y2,y3), then it’s easy to see T(cx+y)=T(cx1+y1,cx2+y2,cx3+y3)=cTx+Ty ( b ) The conditions are a−b=c, the rank of T is 2. ( c ) (a,b,c)=(2t,−4t,−3t),t∈R, the nullity of T is 1.
8. Describe explicitly a linear transformation from R3 into R3 which has as its range the subspace spaned by (1,0,−1) and (1,2,2).
Solution: One possible choice of T can be T(x1,x2,x3)=(x1+x2,2x2+2x3,−x1+2x2+3x3) Since Tϵ1=(1,0,−1),Tϵ2=(1,2,2),Tϵ3=(0,2,3)=Tϵ2−Tϵ1, the condition is satisfied.
9. Let V be the vector space of all n×n matrices over the field F, and let B be a fixed n×n matrix. If
T(A)=AB−BA
verify that T is a linear transformation from V into V.
Solution: Let A,D be any matrix in V, then T(cA+D)=(cA+D)B−B(cA+D)=cAB+DB−cBA−BD=c(AB−BA)+(DB−BD)=cT(A)−T(D)
10. Let 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 into V which is a linear transformation on the above vector space, but which is not a linear transformation on C1, i.e., which is not complex linear.
Solution: Let f be defined as f(a+bi)=a for a+bi∈C (in which a,b∈R), then f is a linear transformation from V into V. To see it’s not a transformation on C1, we have f(1+i)=1, and f(i)=0, but for c=1+i we have f(c(1+i)+i)=f(3i)=0=cf(1+i)+f(i)
11. Let V be the space of n×1 matrices over F and let W be the space of m×1 matrices over F. Let A be a fixed m×n matrix over F and let T be the linear transformation from V into W defined by T(X)=AX. Prove that T is the zero transformation if and only if A is the zero matrix.
Solution: If A is the zero matrix, then obviously T is the zero transformation. Now suppose T is the zero transformation and assume A=0, then at least one aij=0, here 1≤i≤m,1≤j≤n, we let X=aijϵj in V, the j-th coordinate of X is aij and the other coordinates are 0, then the i-th coordinate of AX is aij2=0, thus T(X)=AX=0, a contradiction.
12. Let V be an n-dimensional vector space over the field F and let T be a linear transformation from V into V such that the range and null space of T are identical. Prove that n is even. (Can you give an example of such a linear transformation T)?
Solution: We have rank(T)=nullity(T), since the range of T and the null space of T are equal, thus n=dimV=rank(T)+nullity(T)=2rank(T) so n is even. One example may be T(x1,x2)=(x2,0).
13. Let V be a vector space over the field F and T a linear transformation from V into V. Prove that the following two statements about T are equivalent.
( a ) The intersection of the range of T and the null space of T is the zero subspace of V.
( b ) If T(Tα)=0, then T(α)=0.
Solution: Let A be the range of T and B be the null space of T. First suppose (a) is true, then A∩B={0}, if T(Tα)=0, then Tα∈B, but obviously Tα∈A, thus Tα∈A∩B and Tα=0. Conversely, suppose T(Tα)=0 means Tα=0, and let x∈A∩B, then Tx=0, and there’s β∈V s.t. Tβ=x, so we have T(Tβ)=Tx=0, so Tβ=0=x, thus A∩B={0}.