[理工] 線代

[KAINTS (RUKAWA)]

1.The set of polynomials of degree n is a subsapce
  of the vector space of all polynomials.

  (F): why?

2.The set of all m*n matrices with the usual definitions
  of matrix addition and scalar multiplication is a
  vector space.

  (T):this question does not mention the zero subspace contains
      in the set,so I think the answer is wrong.

                               n  m
3.The vector space Mm*n and L(R ,R ) are isomorphic.

  (T):the definition of isomorphism if the linear transformation
      is one-to-one and onto,that is,the standrd representation
      matrix is invertibe,which the matrix must be squar.
      so I think the answer is wrong.


4.The definite integral is a linear operator on C([a,b]),the
  vector space of cotinuous real-valued functions defined on[a,b].

  (F):why?


5.If a vector space contains a finite linearly dependent set,then
  the vector space is finite-dimesional.

  (F):By GSO ,we can get a orthogonal set,which is also linearly
      independent set,then this set will be finite(becasue a finite
      linearly dependent).
      so I think the answer is true.

6.If T is a linear operator on a vector space with the basis
  B={v1,v2,...v3},then the matrix representation of T with
  respecet to B is the matrix[T(v1) T(v2) ... T(v3)].

  (F):If we replace [T(v1) T(v2) ... T(v3)] by
      [[T(v1)]B  [T(v2)]B  ... [T(v3)]B ],then this question
      will be true?

7.An inner product on a vector space V is linear operator on V.

  (F): why?


8.The indefinite integral can be used to define an inner product
  on P2.

  (F):why?


                                                   n
9.Every subspace of a vector space is a subset of R  for some
  integer n.

  (F): If A is a subspace of B,then A is a subset of B.
       so I think the answer is true.




Def: A linear operator on a finite-dimensional vector space is
     diagonalizable if there is a basis for the vector space
     consisting of eigenvectors of the operator.

我有點不懂這個定義在講什麼= =


問題有點多 感謝板上大大回答

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◆ From: 123.193.7.20
※ 編輯: KAINTS          來自: 123.193.7.20         (08/29 22:52)
推 ILzi:可對角化的定義(算子的部份)
→ ILzi:↑先回答最後一個DEf的問題
推 ddczx:第三題應該是說L這個線性映射可以表示成M的左乘映射，使的每
→ ddczx:一對應的映射均打到同一地方。所以這兩者同構。
→ ddczx:第5題是說向量空間存在有限相依集則向量空間維度為有限，那
→ ddczx:任選無限維向量空間中0向量則為反例
推 ddczx:第9題取多項式空間就不會被包含在R^n的子空間了
→ ddczx:第一題取x^n跟-x^n則相加=0不屬於degree n ,不符合封閉性
可題目有說已經是他的子空間了，我後來想是不是因為這樣
Pn有可能是n+1,n+2,....的子空間
但1,2,...n-1,則不可能，因為題目最後是說all,,所以答案錯
不知道這樣想可以嗎?
→ ddczx:第2題不就是所有m*n矩陣集合的空間，其中當然包含0向量
推 ddczx:第8題不定積分做出來會是多項式，但內積是將兩向量變純量
→ ddczx:第7題 內積:V*V->F，linear operator on V:V->V 所以不一樣
意思有點像，內積是做出一個值;linear operator,則是一個矩陣表示，兩者意思不同，
是這樣嗎?
推 bouwhat:第六題你的附註是正確的by函數矩陣化的定義
推 ILzi:(1)指的是n次多項式，n次多項式並不是向量空間
→ ILzi:(4)要限制在多項式的積分..
為什麼，不太懂ㄟ...

然後可以問一下，inner product space & vector space,
之間有什麼關聯嗎??
※ 編輯: KAINTS          來自: 111.70.142.103       (08/30 10:00)
※ 編輯: KAINTS          來自: 111.70.142.103       (08/30 10:18)
→ ddczx:第一題就是因為題目說是他的子空間但其實不是才錯阿
推 ILzi:cotinuous real-valued functions太多了..e^x也是
→ ghjklgv9:有關聯~單範正交基底就是線性獨立，因為矩陣做單範去除
→ ghjklgv9:掉0，我覺得內積空間好像是種速解，比如說gram的公式
→ ghjklgv9:只要確定生成集合有正交性，就可以用公式速解出某個可以
→ ghjklgv9:向量可以被其他向量做線性組合，這就是第一章的線性系統
→ ghjklgv9:而線性系統用增廣矩陣解AX=B的齊次解，其實也就等於在求
→ ghjklgv9:ker，所以還是有用到前三章的東西，另外還有會用到dim
→ ghjklgv9:rank的看法來說明一些事情，我也才剛上完這邊，有錯誤請
→ ghjklgv9:指點一下，感謝各位前輩
→ ghjklgv9:N次多項式是向量空間的一種吧??子嘉現代四版注意事項3-2