3-1矩陣的運算

3-1 矩陣的運算

3-1矩陣的運算

(甲) 矩陣的基本認識
        請參考講義

(乙) 矩陣的加減法與係數積

(1) A+B = [aij+bij] 
(2) B+A =  [bij+aij] 
(2) A+(B+C) = aij + [bij+cij] 
     (A+B)+C = [aij+bij] + cij 

A - B  = [aij - bij]
B - A = [bij - aij] , but B - A ≠ A- B 
Basically ,   we can conclude the following results : 
1) 矩陣加法滿足交換律. 
2) 矩陣加法滿足結合律
3) 矩陣減法不滿足交換律
4) 矩陣減法不滿足結合律
5) 矩陣其反矩陣存在, 即 A + ( -A) = 0 , 0 是零矩陣 . 
6)  設 A、B、C 都是同階方陣，且 A+B = C，則 A = C−B 且 B = C−A
證明: 
因為  A,B,C 為同階方陣 , 則其加法反元素矩陣存在 , A+(-A) = 0 , B+(-B) = 0 , C+ (-C) = 0 
A+B = C 
(-A) + (A+B) = (-A) + C 
(-A+A) + B = (-A) + C 
0 + B = (-A) + C 
B = C + (-A) 
B = C - A 
同理 , B = C - A 也可以用上面的性質證之. 

[討論]：若 A+B=A+C，則 B = C 恆成立嗎？  
YES. 
A+(-A) = 0  
則  (-A) + A+B = (-A) + A + C 
(-A+ A) + B = (-A+A) + C 
 0 + B = 0 + C 
B = C

但是乘法滿足消去法則嗎 ? 

  




2A = 2[aij] = [2aij] , 3B = 3[bij] = [3bij]  , 2A + 3B = [2aij + 3bij ]

3X - 2B + 3A = 2X - 5C 成立
則 X  = 2B - 3A - 5C , 即求 A,B,C 三個矩陣的乘積之和 .

(1) A - B + 2X = C + 3A + X
      X = C + B - A
(2) 3X + A + B+ C = A - 2B + 4C + X
     2X = -3B + 3C  , X = 3/2 ( C - B)

(1) M = { -3,-2,-1,0,1,2,3,4 } 共8個元素. A ∈ F 3*3  , A = A^t 
   a12,a13,a23,與a21,a31,a32 是一樣的. 
   主對角線 a11,a22,a33 都不變. 所以, 共 6 個位置可以重複選擇, 故有 8⁶  種選擇 . 
(2) skew - Symantec matrix (斜對稱矩陣) , - A = A^t
     同樣是8 個元素,其主對角線必為 0 , 故剩 7 個 元素, 對應 3 個位置可以重複選擇 ,故有 7³  種選擇 . 即剩下的元素 a12, a13, a23 對應 a21 , a31 , a32 . 

(A+B)^t   = A^t + B^t 

A is 2*3 , B is 3*4
AB is 2*4 , but BA is not existed . ( Because the column size of matrix A is not equal to the size of row of  matrix B ) .

(1) 首先 , 計算 A³ 可以得到如下的矩陣 :
  -xy              x - y^2 
  -x^2+xy^2   y^2+1   
   x = 1 , y = 0 

  
  

AB size is 2*2 , (AB)^T  size is 2*2   
B^T  size is 2*3 , A^T size is 3*2 
Calculate  (AB)^T 
B^T A^T  
(AB)^T  = B^T A^T  

     

  

(1) False , A 是  n*n 的方陣 , B 是m*m 的方陣 , 但如果 n = m , A+B 才會有意義 ; 否則 , A+B 無法完成加法運算.
(2) False , AB 成立 ,BA若也存在 , 兩者不一定相同
(3) False , 如上例
(4) True
(5) True

假設 B1 = B2 , 因為 AB2 = B2A  = In ,  
將其代入, AB2 = B2A 得 AB1 = B1A = In 
同理,  也可以得 AB2 = B2A  = In
故 B1 = B2  

AB 有 inverse , 則 (AB)^-1 = B^-1A^-1  這是要證明的事實.
也就是說 , 如果這事實成立  , 則AB 為可逆 , 且 (AB) (B^-1A^-1) = (B^-1A^-1)(AB)  = I 
 (AB) (B^-1A^-1)  = A(BB^-1)A^-1 = AIA^-1 = AA^-1 = I 
  同理 , (B^-1A^-1)(AB) = I       

這可以使用數學歸納法 

n  =  2 , 

(ABA^-1)²  =  (ABA^-1)(ABA^-1)  = AB(A^-1A)BA^-1 = ABIBA^-1 = AB²A^-1 

^{n = k ,  k > 2} 

^(ABA-1)k = AB^kA^-1   holds 

Consider n = k + 1 

AB^kA^-1 (ABA^-1)  = AB^kIA^-1 = AB^kA^-1

A is 4*4 matrix , it can use determent formula to find its inverse . You must use [A|I ] ~ [ I|A^-1] 

( using elementary row operations ) 

Solve it as above

ex 10 . det(A) 不為零. the inverse of  A 可求

ex11 . A無乘法反元素, 即 the inverse of  A 不存在 , 所以  det(A) = 0 , 計算det(A) 令多項式為零求 a

(1) [A|I] ~ [ I|A^-1]
(2) AX = B , X = A^-1B 
(3) XA = B , x  =  ? 

  





5I - A 的反矩陣存在, 即令 X , X 使得 X(5I - A)=(5I - A)X = I . 
5X - XA = I 或 5X - AX = I 
XA = 5X - I 

A²- 5A + 6I  = O 
(A-3I)(A-2I) = O ( Because A is a 2*2 square matrix ) 
A = 3I or A = 2I  
if A= 3I , XA = 5X - I  
X(3I) = 5X - I 
3X = 5X - I 
X = 1/2 I  why ? 
Whether we should use A*A - 5A + 6I calculate them one by one ? 


(丁) 一些特殊矩陣的乘冪

Using mathematics induction can prove that the fact of Aⁿ   

如果找出  B⁴ ,  B⁵  之後就可用歸納假設的方式 假設Bⁿ = 0 , n > 3 再使用mathematics induction 證明成立. 同理A^{n 亦同 .} 

D = PA P^{-1 
Because the inverse of P exists , then above expression can be expressed as below}  

 P^-1DP = A

Dⁿ = (PAP^-1)(PA P^-1)(PA P^-1) ^...(PA P^-1) (PA P^-1)  

      = PA(P^-1P)A(P^-1P) ... A(P^-1P)AP^-1 

       = PAⁿP^-1 

 Aⁿ = P^-1Dⁿ P 

先計算2,3次方,分析其結果,在假設高次方. 即可求出

main diagonal matrix , A  , A², A³   , ... , Aⁿ    can be written as below :
Aⁿ = 2^n      0         0 
         0       (-1)^n    0  
         0        0          3^n 

then you may use above expression to calculate A^2 , A^3 . 

To prove the fact , we can calculate A^3  ( because it is the induction basis ) , and using maths induction to prove it.

Because A is an upper triangle matrix . According to the property of upper matrix ,

觀察 A的內容 A = 2I  + B  , B is a upper triangle matrix . So we can assume k = 2 . 

B^2  =   0 0 3
             0 0 0
             0 0 0
B^3 = 0

So A^5 = (kI+ B)^5  = (2I + B)^5
觀察  B^3 = 0 , 所以 從B的次方有大於 or 等於 3 的項目其乘積項皆為零矩陣,故不考慮.
((5,5)2^5)I^5B^0 + (5,1)2^4I^4B + (5,2)2^3I^3B^2
32 , 5*16*3 = 240  , 10*8*3 = 240
所以是 240 答案有誤

(1) using the formula
(2) Diagonalization , D = P^-1AP , D is a main diagonal matrix , D = d1   0     0
                                                                                                              0     d2   0
                                                                                                              0     0     d3

 

(3) calculate A^2, A^3 and obverse their result and assume a conclusion , finally using mathematical induction  to prove it.

How to resolve this question quickly ? 

 To evaluate Aⁿ  , we should calculate A²  A³ A⁴ ,  ... then discover its result and conclude its form.
A = cosθ    sinθ
     sinθ   -cosθ 
A²  =   I2 

A³  = A² ×A  = A 
A⁴  = A³ ×A  = A×A = A²  
... 
So we can conclude a result as below :  
Aⁿ  = A ,  n is odd
         I2 ,   n is even

綜合練習

    

2(X+B-A) = - (2B-3X) + A 

2X + 2B - 2A = -2B + 3X + A 

-X = - 4B + 2A + A 

-X = - 4B + 3A 

       





cos(A+B) = cosAcosB - sinAsinB 
cos(θ+θ) = cos(2θ)  =   cos²θ - sin²θ 
A = sinθ , B = cosθ  
A+B = 1/2  
2AB = b 
B²  - A² = a
(B-A)(B+A) = a
B - A = c  
c/2 = a 
c = 2a  
B²  + A² = 1/4 - b 
Since B²  + A²  = cos²θ + sin²θ = 1 , 1 = 1/4 -b , b = -3/4  代回 AB = - 3/ 8 . 
2B²  = a + 1 , B² = (a + 1)/2

(A) False 
(B) False 
(C) False  
(D) True 
(E) False  

(A) True.   det(AB) = det(A)det(B) = det(B)det(A) = det(BA) = det (O) = 0  
(B) True. det(AB)= det (A)det(B) = det(B)det(A) = det(BA) = 1 
(C) (A+I)(A-I) = A²- A + A - I = A² - I 
(D) 若 令 A² - I = O , (A+I)(A-I) = O , A = I or A = - I . 
(E) 若  AB = O , For counterexample ,  A ≠ O , A = 3   6     B = 2       3
                                                                                                       ^{-2   -4}            -1    - 3/2

   



(A) det(A+B) = det(A) + det(B) is false. For example , A =  1    -1  ,  B = -1     1 
                                                                                                 2     2            -2    -2 
det(A) ≠ 0 , det(B) ≠ 0 , but det(A+B) = 0 

(B) False det(kA) = k^ndet(A) 
(C) det(A) ≠ 0 , the inverse of  A exists . AB = AC , applies the inverse of A to two sides of equal sign and get B = C . 
(D) True. Prove it by yourself .  
(E) FALSE  BECAUSE AB ≠ BA 

(A) A^-1存在 ,  AA^-1 = A^-1A = I . det(I) = det(A^-1A) = det(A)det(A^-1) = 1 
(B) true 
(C) (AB)² = (AB)(AB) = A(BA)B ≠ A(AB)B ( AB 不一定等於 BA ) 
(D) 若 A ≠ O 且 B ≠ O 則 AB ≠ O 錯, 可看上例
(E) AX = C 恰有一組解 , 即 the inverse of A 存在 ,  det(A) ≠ 0  

      




(a) 套公式算較快
(b) 同上
(c)  先求 A² , A³  , A⁴,  ... 觀察其內容即可歸納 Aⁿ 然後 ,  使用數學歸納法證之. 
(d) B2 = I2  ,  B3 = I2 ×B = B , B4 = B2× B2 = I2 ×I2  = I2  , ...  
      Bn  =  B   , n is odd
                 I2   , n is even

(e)  AB100A^-1 = AIA^-1= AA^-1 = I

(1) A = A^-1 , A² = I , A³ = A ,  ...   ; B^-1 也存在 ,
     B3 = - I
     B4 = - B
     B5 = B4 × B = - B × B = - B² 
     B6 = B5 × B =  - B² × B  =  I
     B7 = B6 × B = B
     B8 = B × B = B^{2 
     B9 = B × B2 
       ...} 
     Bn  = (-1)nI    , n = 3k , k = 1,2,3, ...
              (-1)nB   , n = 2k +3 , k = 0,1,2,3, ..
              (-1)nB2 , n = 3k -1 , k = 1,2,3, ..
                 
     

(2) True
(3) False , - A = A
(4) the inverse of A exists , apply A^-1 to two sides of  equal sign and obtain B¹² = A6   , is it true ?
      B¹²   = I , A6 =  A3  ×  A3 = A  × A = I
(5) (ABA)15 = (ABA)(ABA)(ABA) ...(ABA)(ABA)  = AB(AA)B(AA)B(AA) ...B(AA)BA = ABB...BA = AB15A = A(-B)A = - ABA

先將A乘開 , 然後在陸陸續續乘前面的. 求 a , b, c,d 即可.

X+Y = A , X -Y  = B
2X = A + B
X  = 1/2 [A+B] = 1    2   代回 Y =   1     0  
                             3    4                   -1      1

x^2 -2x - 3I = O , x(x-2I) - 3I = O , x(x-2I) = 3I 因為 A滿足左式. 所以, A(A-2I) = 3I

A[1/3(A-2I)] = I ,  A^-1 = 1/3(A-2I) .  

計算這個首先需要先觀察, 我們將第一列乘 cosθ , 得 cos2θ    -sinθcosθ    0 
第二列乘上sinθ ,得 sin2θ     cosθsinθ    0 接著 row 1 add to row 2 則得  cos2θ +  sin2θ      0      0  
等於 1     0      0 , row 1 及 row 2 互換. ( row change )

(a) [A|I] ~ [I|A^-1]
(b) X = A^-1 [ -1 2 1 ]^t
(c) Since  A^-1 exists , A can be reduced to an elementary row operation matrix , det(5A) = 125 det(I) = 125

(a) A is invertible iff  det(A) ≠ 0 , x's condition can be found . 
(b) x = 4 代入, [A|I] ~[ I| A^-1] 

法一 : 
if A hasn't the inverse , det(A) = 0 , 
法二 : 
[A|I] ~[I|A^-1] 求出 A^-1 再比較 k值 

AX = B , B = [ 9    3    3 ]^{t   
Let X ' =} [ a    b   c ]^{t  ,  Y =} [ 1    0    1 ]<sup>t  

AX'  = Y</sup> 
A^-1  Y = X' 

(a) 乘一下
(b) PAnP-1  = Dn  
      PDnP-1  = An    

Mathematical induction
Induction basis :
 n = 1 , trivial true .
Inductive steps :
n = k , k > 1
A^k holds
Consider  n  = k +1 , A^k * A = a^(k+1)      (k+1)a^k
                                                  0                         a^(k+1)

pf:

if
expand (A+I)(A-I) = AA - A + A - I = AA - I = 0 if AA = I
only if
AA = I
AA - I = 0
AA - A + A - I = 0
A(A-I)+(A-I) = 0
A(A-I)+I(A-I) = 0
(A+I)(A-I) =0














































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