[線代] ║A║=║A^*║ 證明從定義下手
Let A€M_n(F):={square matrix consisting of elements in field F=R or C}
and ║A║:= sup{│Av│：│v│=1, v€F^n}
Then ║A║=║A^*║, where A^* is conjugate transpose of A
這個如果用"A(A^*)與(A^*)A有相同的特徵方程式"這個定理
就可以用║A║^2 = largest eigenvalue of A^*A 來秒殺
一般都是這樣證的嗎?? 會不會殺雞用牛刀了@@??
感覺有從定義做的可能性，即觀察<Av,Av>之類的
謝謝
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