Prove a parte (d) do Teorema 1.4.8.

Passo 1

O item (d) diz que (k A)^T = k A^T , vamos encontrar os dois lados da equação e comparar

A = [[[a_11,a_12,cdots,a_1 j],[a_21,a_22,cdots,a_2 j],[vdots,vdots,ddots,,vdots],[a_i 1,a_i 2,cdots,a_i j]]] => A^T = [[[a_11,a_21,cdots,a_i 1],[a_12,a_22,cdots,a_i 2],[vdots,vdots,ddots,,vdots],[a_1 j,a_2 j,cdots,a_i j]]] =>

=> k A^T = [[[k a_11,k a_21,cdots,k,a_i 1],[k,a_12,k,a_22,cdots,k,a_i 2],[vdots,vdots,ddots,,vdots],[k,a_1 j,k,a_2 j,cdots,k a_i j]]]

Agora o outro lado

A = [[[a_11,a_12,cdots,a_1 j],[a_21,a_22,cdots,a_2 j],[vdots,vdots,ddots,,vdots],[a_i 1,a_i 2,cdots,a_i j]]] => k A = [[[k a_11,k,a_12,cdots,k a_1 j],[k a_21,k a_22,cdots,k,a_2 j],[vdots,vdots,ddots,,vdots],[k,a_i 1,k,a_i 2,cdots,k,a_i j]]] =>

=> (k A)^T = [[[k a_11,k a_21,cdots,k,a_i 1],[k,a_12,k,a_22,cdots,k,a_i 2],[vdots,vdots,ddots,,vdots],[k,a_1 j,k,a_2 j,cdots,k a_i j]]]

Que são EXATAMENTE IGUAL!!!!

Resposta

(k A)^T = [[[k a_11,k a_21,cdots,k,a_i 1],[k,a_12,k,a_22,cdots,k,a_i 2],[vdots,vdots,ddots,,vdots],[k,a_1 j,k,a_2 j,cdots,k a_i j]]] = k A^T

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