gaussianeliminationisstablefortheinverseofadiagonallydominantmatrix-外文文献(编辑修改稿)

gaussianeliminationisstablefortheinverseofadiagonallydominantmatrix-外文文献(编辑修改稿)内容摘要：

e, it is denoted as A=A( )。 where(7) =f1。 :::。 kg:Lemma 3. It holds that(8) (A=A( )) 1=B( 0)。 where B =A 1:License or copyright restrictions may apply to redistribution。 see GAUSSIAN ELIMINATION 655This is a wellknown relation (see, for example, [4, Sec. ]).Let a(k)ij(i。 j =k +1。 :::。 n ) be the entries of the Schur plement A=A( )。 being the index set in (7). The quantity(9) n(A)=maxi。 j。 kja(k)ijjmaxi。 jjaijjis called the growth factor for A:The properties of . matrices related to Gaussian elimination are widely known.We state those we need in Section 3 in the lemma below.Lemma 4. Let B2 Mn(C) be a . matrix with the row dominance factors i(see (3)). Then:(1) Gaussian elimination is applicable toB under any diagonal pivoting order.(2) The diagonal dominance property is inherited by active submatrices. Inother words, each Schur plement B=B( ) is also a . matrix. Moreover, for each i。 the row dominance factor 0ifor B=B( ) does not exceedthe corresponding factor ifor B (assuming that the original row indicesof B remain \attached to the rows in B=B( )).3. The main resultWe now proveTheorem 1. Let A2Mn(C) be a nonsingular matrix such that B =A 1is a .matrix with the dominance factor (see (4)). Then(10) n(A) 1+:Proof. By Lemma 2,a11is the entry with the largest modulus in the rst column.Thus,a11can be taken as the pivot for the rst step of elimination. Setting =f1g。 we see from (8) that A=A( ) has the . matrix B( 0) as its inverse. Hence, a(1)22is the entry with the largest modulus in the rst column of A=A( )a n dc a nb etaken as the pivot for the second step. Continuing in this way, we conclude that nopermutations are needed to perform Gaussian elimination (GE) for A: Moreover,GE with no pivoting as applied to A is the same as GE with partial pivoting.In fact, relation (6) means that the entry with the largest modu。