Question

线性代数证明题 高手入！！

Let T : Rm → Rn be a linear transformation.
(1) Let B = (v1,v2, . . . ,vm) be a basis of Rm. Let k be an integer such that 0 ≤ k ≤ m and the first k vectors of B form a basis of ker(T). Prove that T(vk+1), . . . , T(vm) form a basis of im(T).
(2) Does every basis of Rm contain a basis of ker(T)?
(3) Does there always exist a basis B of Rm that contains a basis of ker(T)?

Answer 1

只给提示，不给答案，不要问我为什么，因为任性：
1，考虑im(T)中任意元素的原像可由v1,v2,...,vm线性表出，所以im(T)的任意元素可由T(v1),T(v2),...,T(vm)线性表出，由v1,v2,...,vk为ker(T)元素，可知结论成立
2，反例Rm=L(v1,v2,v3)；ker(T)=L(v1,v2)；B=(v1+v3,v2+v3,v3),显然B是Rm的一组基，但B不包含ker(T)的基
3，正确，扩充基定理

Answer 2

(1) It is easy to see that Im(T) is a vector space (or linear space) ,for T: Rm->Im(T) is surjective, It is no doubt T(vk+1), . . . , T(vm) can span Im(T) except 0 ——just verify T(vk+1), . . . , T(vm) are linearly independent——if not, there'll be xk+1,....,xm s.t. T(∑xivi) = 0,i>=k+1 and <=m,so ∑xivi belongs to kerT which is wrong for B contains only linearly independent vectors ;
(2) Of course not----let m = n = 2,Rm = Rn = span(v1,v2),T is a projection which send v2 to 0 and Tv1 = v1, it is easy to verify T is linear,and (v1,v1+v2) is a basis of Rm,but kerT = span(v2),which isn't contained in (v1,v1+v2);
(3)It is right——kerT is clearly a linear space and it must have a basis B,then any basis C of Rm/kerT can form a basis of Rm combined with the basis of kerT,Rm is isomorphic to BxC,'x' is the product and you can write it as a direct sum.