Closed Sketh closed 5 years ago
因为 (q ∗ ) ∗ = [s, −(−v)] = [s, v] = q， q ∗ q = (q ∗ )(q ∗ ) ∗ = ∥q ∗ ∥ 2 = s 2 + x 2 + y 2 + z 2 = ∥q∥ 2 = qq∗ 所以我们得到，q ∗ q = qq∗．这个特殊的乘法是遵守交换律的．
請問為甚麼在q = (q ∗ )(q ∗ ) ∗ = ∥q ∗ ∥ 2 這裡的時候最後結果有一個共軛符號消失了
q = (q ∗ )(q ∗ ) ∗ = ∥q ∗ ∥ 2
因为 (q ∗ ) ∗ = [s, −(−v)] = [s, v] = q， q ∗ q = (q ∗ )(q ∗ ) ∗ = ∥q ∗ ∥ 2 = s 2 + x 2 + y 2 + z 2 = ∥q∥ 2 = qq∗ 所以我们得到，q ∗ q = qq∗．这个特殊的乘法是遵守交换律的．請問為甚麼在
q = (q ∗ )(q ∗ ) ∗ = ∥q ∗ ∥ 2這裡的時候最後結果有一個共軛符號消失了