Enunciado

Prove o Teorema 5.3.2

Passo 1

O Teorema 5.3.2 diz que:

  1. overset(-)(overset(-)(A)) = A
  2. overset(-)((A^T)) = (A)^T
  3. overset(-)(A B) = overset(-)(A) overset(-)(B)

Para isso vamos usar duas matrizes:

A_n k = [[[a_11,+,i,b_11,a_12,+,i,b_12,cdots,a_1 k,+,i,b_1 k],[a_21,+,i,b_21,a_22,+,i,b_22,cdots,a_2 k,+,i,b_2 k],[vdots,vdots,cdots,vdots],[a_n 1,+,i,b_n 1,a_n 2,+,i,b_n 2,cdots,a_n k,+,i,b_n k]]]

B_k m = [[[c_11,+,i,d_11,c_12,+,i,d_12,cdots,c_1 m,+,i,d_1 m],[c_21,+,i,d_21,c_22,+,i,d_22,cdots,c_2 m,+,i,d_2 m],[vdots,vdots,cdots,vdots],[c_k 1,+,i,d_k 1,c_k 2,+,i,d_k 2,cdots,c_k m,+,i,d_k m]]]

Passo 2

Vamos provar o Teorema 5.3.2a)

Dada uma matriz no formato:

A = [[[a_11,+,i,b_11,a_12,+,i,b_12,cdots,a_1 k,+,i,b_1 k],[a_21,+,i,b_21,a_22,+,i,b_22,cdots,a_2 k,+,i,b_2 k],[vdots,vdots,cdots,vdots],[a_n 1,+,i,b_n 1,a_n 2,+,i,b_n 2,cdots,a_n k,+,i,b_n k]]]

Vamos aplicar um conjugado em em A, vamos trocar os sinais que ficam em frente aos i ’s

overset(-)(A) = [[[a_11,-,i,b_11,a_12,-,i,b_12,cdots,a_1 k,-,i,b_1 k],[a_21,-,i,b_21,a_22,-,i,b_22,cdots,a_2 k,-,i,b_2 k],[vdots,vdots,ddots,vdots],[a_n 1,-,i,b_n 1,a_n 2,-,i,b_n 2,cdots,a_n k,-,i,b_n k]]]

E vamos aplicar um conjugado mais uma vez, trocando novamente os sinais:

overset(-)(overset(-)(A)) = [[[a_11,+,i,b_11,a_12,+,i,b_12,cdots,a_1 k,+,i,b_1 k],[a_21,+,i,b_21,a_22,+,i,b_22,cdots,a_2 k,+,i,b_2 k],[vdots,vdots,cdots,vdots],[a_n 1,+,i,b_n 1,a_n 2,+,i,b_n 2,cdots,a_n k,+,i,b_n k]]]

Ou seja:

overset(-)(overset(-)(A)) = A

Passo 3

Vamos provar o Teorema 5.3.2b)

Primeiro vamos testar o conjugado da transposta:   overset(-)((A^T))

A = [[[a_11,+,i,b_11,a_12,+,i,b_12,cdots,a_1 k,+,i,b_1 k],[a_21,+,i,b_21,a_22,+,i,b_22,cdots,a_2 k,+,i,b_2 k],[vdots,vdots,cdots,vdots],[a_n 1,+,i,b_n 1,a_n 2,+,i,b_n 2,cdots,a_n k,+,i,b_n k]]]

A t r a s p o s t a d e s t a m a t r i z é

A^T = [[[a_11,+,i,b_11,a_21,+,i,b_21,cdots,a_n 1,+,i,b_n 1],[a_12,+,i,b_12,a_22,+,i,b_22,cdots,a_n 2,+,i,b_n 2],[vdots,vdots,vdots,vdots],[a_1 k,+,i,b_1 k,a_2 k,+,i,b_2 k,cdots,a_n k,+,i,b_n k]]]

E a p l i c a n d o o c o n j u g a d o :

overset(-)(( A^T )) = [[[a_11,-,i,b_11,a_21,-,i,b_21,cdots,a_n 1,-,i,b_n 1],[a_12,-,i,b_12,a_22,-,i,b_22,cdots,a_n 2,-,i,b_n 2],[vdots,vdots,vdots,vdots],[a_1 k,-,i,b_1 k,a_2 k,-,i,b_2 k,cdots,a_n k,-,i,b_n k]]]

Agora vamos testar a transposta do conjugado: ( overset(-)(A) )^T

A = [[[a_11,+,i,b_11,a_12,+,i,b_12,cdots,a_1 k,+,i,b_1 k],[a_21,+,i,b_21,a_22,+,i,b_22,cdots,a_2 k,+,i,b_2 k],[vdots,vdots,cdots,vdots],[a_n 1,+,i,b_n 1,a_n 2,+,i,b_n 2,cdots,a_n k,+,i,b_n k]]]

A p l i c a n d o o c o n j u g a d o :

overset(-)(A) = [[[a_11,-,i,b_11,a_12,-,i,b_12,cdots,a_1 k,-,i,b_1 k],[a_21,-,i,b_21,a_22,-,i,b_22,cdots,a_2 k,-,i,b_2 k],[vdots,vdots,ddots,vdots],[a_n 1,-,i,b_n 1,a_n 2,-,i,b_n 2,cdots,a_n k,-,i,b_n k]]]

E a p l i c a n d o a t r a n s p o s i ç ã o :

(overset(-)(A))^T = [[[a_11,-,i,b_11,a_21,-,i,b_21,cdots,a_n 1,-,i,b_n 1],[a_12,-,i,b_12,a_22,-,i,b_22,cdots,a_n 2,-,i,b_n 2],[vdots,vdots,vdots,vdots],[a_1 k,-,i,b_1 k,a_2 k,-,i,b_2 k,cdots,a_n k,-,i,b_n k]]]

Comparando os dois resultados, temos que:

(overset(-)(A))^T = overset(-)(( A^T ))

Passo 4

Agora vamos provar o Teorema 5.3.2c)

Primeiro vamos calcular o conjugado do produto   overset(-)(A B) :

A B = [[[a_11,+,i,b_11,cdots,a_1 k,+,i,b_1 k],[vdots,ddots,vdots],[a_n 1,+,i,b_n 1,cdots,a_n k,+,i,b_n k]]] [[[c_11,+,i,d_11,cdots,c_1 m,+,i,d_1 m],[vdots,ddots,vdots],[c_k 1,+,i,d_k 1,cdots,c_k m,+,i,d_k m]]]

Vamos aplicar o produto matricial:

A B = [[[sum_1^k ( a_1 j + i b_1 j ) ( c_j 1 + i d_j 1 ),cdots,sum_1^k ( a_1 j + i b_1 j ) ( c_j m + i d_j m )],[vdots,ddots,vdots],[sum_1^k ( a_n j + i b_n j ) ( c_j 1 + i d_j 1 ),cdots,sum_1^k ( a_n j + i b_n j ) ( c_j m + i d_j m )]]]

E agora aplicamos o conjugado:

overset(-)(A B) = [[[sum_1^k ( a_1 j - i b_1 j ) ( c_j 1 - i d_j 1 ),cdots,sum_1^k ( a_1 j - i b_1 j ) ( c_j m - i d_j m )],[vdots,ddots,vdots],[sum_1^k ( a_n j - i b_n j ) ( c_j 1 - i d_j 1 ),cdots,sum_1^k ( a_n j - i b_n j ) ( c_j m - i d_j m )]]]

Podemos separar agora em duas matrizes:

overset(-)(A B) = [[[a_11,-,i,b_11,cdots,a_1 k,-,i,b_1 k],[vdots,ddots,vdots],[a_n 1,-,i,b_n 1,cdots,a_n k,-,i,b_n k]]] [[[c_11,-,i,d_11,cdots,c_1 m,-,i,d_1 m],[vdots,ddots,vdots],[c_k 1,-,i,d_k 1,cdots,c_k m,-,i,d_k m]]]

E essas matrizes que separamos são os conjugados de A e B :

overset(-)(A B) = overset(-)(A) * overset(-)(B)

Resposta

Ei, a resposta está no passo a passo :)

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