例2计算矩阵A 的各种范数 解(1)4|=max∑|an|=max{+1,2+3}=5, I≤n (2川A=max∑|an|=max1+2,1+3}=4, ≤ (3AA「55 元-5-5 九I-AA|= 510 5元-10 1≈13.091,12≈1.910, x2-154+25=0, Al2=√am(A4)≈√3.091=3618, (4=、∑=√+1++1=15 l,例2 计算矩阵 的各种范数。 解 1 1 2 3 A   − =     1 1 1 (2) || || max | | n ij j n i A a   = =  2 max  = || || ( ) A A A   1 1 (1) || || max | | n ij i n j A a    = =  = + + = max{1 2,1 3} 4, 5 5 (3) , 5 10 A A    =     2 5 5 | | 5 10 15 25 0, I A A      − − − =  − − = − + = 1 2     13.091, 1.910,  = 13.091 3.618, = + + = max{1 1, 2 3} 5, 2 , 1 (4) || || n F ij i j A a = =  = + + + = 1 1 4 9 15