×

Loading...
Ad by
  • 最优利率和cashback可以申请特批,好信用好收入offer更好。请点链接扫码加微信咨询,Scotiabank -- Nick Zhang 6478812600。
Ad by
  • 最优利率和cashback可以申请特批,好信用好收入offer更好。请点链接扫码加微信咨询,Scotiabank -- Nick Zhang 6478812600。

You owe me big time

1.
The following makes assumption that f and g are both linear......

f o g (v) = f(g(v))

if f(g(v)) is one to one
>> f(g(v)) = 0 if g(v) = 0 and there is only 1 v such that g(v) = 0
>> v = 0 only.
Therefore, g is one-to-one because g(v) = 0 iff v=0

2.
This is actually a corollary. Again, the assumption is f is liinear.

if f is one-to-one
>> ker(f) = {0}
>> Nullity(f) + Rank(f) = dim(A)
>> Rank(f) = dim(A) = dim(B)
>> Range(f) is a subset of A with the same dimension.
>> Range(f) = B
>> we prove if it's one-to-one, then it's onto

if f is onto
>>Range(f) = B
>>Rank(f) = dim(B)
>>Nullity(f) + dim(B) = dim(A) = dim(B)
>>Nullity(f) = 0. ie. ker(f) = {0}
>>f is one-to-one

Hence, this prove if and only if.
Report

Replies, comments and Discussions:

  • 工作学习 / 求学深造 / help, help, help 谁能帮助俺解2个离散数学的题---
    1.If f and f o g are one-to-one, does it follow that g is one-to-one? Justify your answer.

    2.Suppose that f is a function from A to B, where a and B are finite sets with |A| = |B|. Show that f is one-to –one if and only if it is onto.
    • 要在上学期我就能帮你做了,现在书都卖了,也忘的差不多了。什么时候交?晚几天倒是能做出来。帮你UP吧。
      • 明天早上:((
        • 真同情..... :-D
    • 虽然知道自己学过但肯定不会做了,但还是忍不住进来看看,不看不知道
      一看才知道,自己题目也看不懂,咳,我这英文该回去还给老师了
      • 不管怎么说,谢谢好心肠的MM~~~
        • 你不能这么害我啊, 我不能帮你,你也不能把我变性了 :((((((((
    • what's the meaning of "f o g" in 1st question ?
      • 复合函数 f(g(x))
        • 可惜俺E文不行,不知道怎么讲~~~给你一个中文参考。§8 部分
          • 俺对概念还是有点明白的。就是俺对函数从小就怕,尤其是证明这个那个,长大了学完了以为再也不要了,结果现在还是要学,命真苦呀:(((
            • 套用一下定义定理就好呀
    • 大学时就离散数学枯燥无味难学,考试时有十几个人booking俺周围的位子,其中包括几位恐龙mm。俺掂量再三,小心地选择了6个mm(锉子里头拔高个),2个gg(因为他们是我打勾级时的联邦),结果,kao
      有一个mm的成绩比我还高。原来,她收集了3份答案,少数服从多数,择优选择。
      不过,现在那点底都丢到爪哇国了,连题都看不懂了,惨!
      >
      >
      >
      >
      >
      >
      我是说雨雨mm好惨!
      • 想起了那个笑话,说一个学生用掷骰子/硬币的办法做选择题,每一题都多掷好几遍,老师问他为什么,他回答是
        我总得验算一次啊
    • You owe me big time
      1.
      The following makes assumption that f and g are both linear......

      f o g (v) = f(g(v))

      if f(g(v)) is one to one
      >> f(g(v)) = 0 if g(v) = 0 and there is only 1 v such that g(v) = 0
      >> v = 0 only.
      Therefore, g is one-to-one because g(v) = 0 iff v=0

      2.
      This is actually a corollary. Again, the assumption is f is liinear.

      if f is one-to-one
      >> ker(f) = {0}
      >> Nullity(f) + Rank(f) = dim(A)
      >> Rank(f) = dim(A) = dim(B)
      >> Range(f) is a subset of A with the same dimension.
      >> Range(f) = B
      >> we prove if it's one-to-one, then it's onto

      if f is onto
      >>Range(f) = B
      >>Rank(f) = dim(B)
      >>Nullity(f) + dim(B) = dim(A) = dim(B)
      >>Nullity(f) = 0. ie. ker(f) = {0}
      >>f is one-to-one

      Hence, this prove if and only if.
      • "f(g(v)) = 0 if g(v) = 0" ?
        • Did you even read the question? It's given f is one to one.
          • "function such that each element of the range has a unique preimage are called one-to-one" this is df. of one-to-one. It's impossible to get "f(g(v)) = 0 if g(v) = 0 " by assume "v=0".
          • c/m:f(x) = x + 1, g(x)=x, x=0 g(x)=0, f(g(0))=1
            • yo....just to refresh your memory....we are talking about vector spaces...1 could be 0 vector under a vector space... ok??
              • ....right....but still think your pf. has problem...sigh, forget almost of math, don't like those courses.
                • I took them 2 terms ago and forgot all of them as well.. copied straight out from my notes.
                  • suggest rainrain directly uses your ans, let's wait and see if it's right :P...if it's same question as your notes, maybe I am wrong...but I think there should be another way to proof that.....
                    • I made a comment below...... My proof is about map and transformation of vector spaces. But her questions are about discrete mathematics, maybe it's the same.
                    • The whole thing is like this if you are not convinced this is right under linear algebra.
                      Suppose L: V --> W is a linear map and V and W are vector spaces.

                      Because it's linear, then there exists.
                      L(0v) = 0w where 0v is the zero vector under vector space V
                      0w is the zero vector under vector space W
                      This is a property of linear map....

                      We are given L is one to one....

                      Therefore, 0v is the only solution.....
                      • I got rainrain's questions from my friend's discrete math textbook, it doesn't use such algebra way to solve...just few sentense...
                        • okee..
    • Isn't this called Linear Algebra? 线性代数?Instead of 离散数学?
      • should be linear algebra
      • thanks, many thanks, it's discrete math.
        还有一个f is a function from A to B, let S and T be the subsets of B. Show that f -1(S U T) = f -1(S) U f-1 (T)
        f-1是说的逆函数,俺不会打那个小-1 :(

        俺无以为报,俺只能在任何话题上无条件支持你:(
        • Sorry. can't find it in my note.
    • 我真同情你. 实在不行,找个同学问问吧.
    • Ok... I have to remind you that proof I gave you is valid under Linear Algebra where we are dealing vector spaces and different type of maps. I'm not sure whether it's the same under discrete mathematics.
    • 1.If f and f o g are one-to-one, does it follow that g is one-to-one? Justify your answer.
      Here, we assume that G is a function from some set A to another set B, while F is a function from B to some set C. Consider two distinct points, a,a' of A. Since F o G is one-to-one, we know that c= F(G(a)) is distinct from c' = F(G(a')). Of course the only way that this could happen is if G(a) is distinct from G(a'), i.e. just think about it if they were the same. Since a and a' is an arbitrary pair of distinct elements of A, this shows that G is one-to-one.
      • Long time no see...How's the weather in Edmonton? We have a record high in southern Ontario...28 degree yesterday and today... crazy!!!
        • 嘿嘿,看看我的答案对不对。题目反正是对的。我们这里现在早晨只有3度。不过屋里很暖和。
          • I don't know ar... I thought her question is about linear algebra since I didn't know 离散数学 is discrete math....... I haven't and won't be taking any discrete math in my life....
      • 2.Suppose that f is a function from A to B, where a and B are finite sets with |A| = |B|. Show that f is one-to –one if and only if it is onto.
        Let n = |A|. Thus A = {x1,x2,x3, ¼,xn}, where all elements are distinct.

        Assume that f is a one-to-one function. Then for any integer j we have f(xi) &sup1; f(xj) whenever 0 &pound; i < j. This implies that |{f(x1),f(x2),f(x3), &frac14;,f(xn)}| = n. Since {f(x1),f(x2),f(x3), &frac14;,f(xn)} &Iacute; B and |{f(x1),f(x2),f(x3), &frac14;,f(xn)}| = |B|, we can conclude that the range of f is B. Therefore, f is an onto function. Assume now that f is onto but it is not a one-to-one function. Then there exist integers i &sup1; j such that f(xi) = f(xj). Thus, |{f(x1),f(x2),f(x3), &frac14;,f(xn)}| < n and since |B| = n, there must be an element of B so that no element of A is mapped on it by f. Thus, f is not onto, a contradiction.