证:1):a=0或a=1 Gerschgorin定理 2): Ax=/x|= 2 aij l ≠L ∑|l1x;1=∑|a;(a;21x;p 丿≠L j≠i s|20/gy1a1 f1∑(ax;1a)}=a j=1 J= J≠返回 证: 1):  = 0或 = 1 2): Ax = x Gerschgorin定理 | | | | | | 1 − =   = n j i j  aii xi aij x j | | | | 1    = n j i j aij x j | | (| | | |) 1 1 =   = − n j i j aij aij x j       − −  =  = −    1/ (1 ) 1 1 1 1/ 1 [ (| | ) ] [ (| | | |) ] n j i j n j i j i j i j j a a x