卜長(zhǎng)江,劉廣峰,白淑艷
(哈爾濱工程大學(xué)理學(xué)院,黑龍江哈爾濱150001)
矩陣的Drazin逆表示不僅在矩陣代數(shù)中有重要意義,而且在奇異差分方程、廣義系統(tǒng)、馬爾科夫鏈和迭代法等領(lǐng)域中都有十分重要的應(yīng)用[1-5].尤其在奇異線性微分方程的解表示中[6],矩陣的Drazin逆理論更是不可或缺的工具.1979年,以二階奇異系統(tǒng)為背景,Campbell等[1]提出一個(gè)2×2分塊矩陣Drazin逆表達(dá)式的open問(wèn)題.1983年,Campbell[2]再次提出類似的open問(wèn)題,引起學(xué)者們極大的關(guān)注,但這些問(wèn)題至今未能完全解決.近年來(lái)一些學(xué)者在分塊矩陣的子塊滿足一定條件下給出其Drazin逆表達(dá)式[7-14].在分塊矩陣的廣義Schur補(bǔ)為零或者可逆的條件下給出其Drazin逆表達(dá)式的問(wèn)題是近幾年的熱點(diǎn)問(wèn)題[7-10].本文在分塊矩陣的廣義Schur補(bǔ)為零且子塊滿足相應(yīng)條件下給出了其Drazin表達(dá)式,推廣了 Hartwing[9]及 Martinez-Serrano[10]給出的若干結(jié)果.
為了得到本文的主要結(jié)果首先給出如下引理.引理 1[10]設(shè) P,Q∈Cn×n,若 P2Q=O,Q2=O且PQ是k次冪零矩陣,則
其中:k=ind(PQ),t=ind(P2).
引理 2[10]P,Q∈Cn×n,若 P2Q=O,Q2=O,QPQ=O且(PQ)2=O,則(P+Q)D=PD+Q(PD)2+PQ(PD)3.
該文主要是得到下面形式的2×2分塊矩陣Drazin逆表達(dá)式
式中:A和D是方陣,設(shè)M的廣義Schur補(bǔ)S=DCADB等于零.
證明:由于M=PNP-1,令N=P1+P2,
所以,由引理1得
因?yàn)镸D=PNDP-1,
下面給出定理1的推論,此推論恰好是文獻(xiàn)[10]的結(jié)果.
推論 設(shè)分塊矩陣M如同式(1).若BCAπ是r次冪零矩陣,ABCAπ=O,則
其中:
文獻(xiàn)[9]在 CAπB=O,AAπB=O 條件下給出分塊矩陣M的Drazin逆表達(dá)式.本文的定理2在CAπBC=O且AAπBC=O的條件下給出M 的的Drazin逆表達(dá)式.顯然,CAπBC=O和AAπBC=O是CAπB=O和AAπB=O成立的必要條件.
定理2 設(shè)分塊矩陣M的形式如同式(1).若CAπBC=O,AAπBC=O,則
其中:B11、B12、Z 如定理中所示.
通過(guò)與定理1和定理2類似的證明方法分別給出定理3和定理4.
證明:證明過(guò)程與定理2類似,略.
定理4 設(shè)分塊矩陣M的形式如同式(1).若AπBC 是 p次冪零矩陣,(A+ADBC)AπBC=O,則
其中:
證明:證明過(guò)程與定理1類似,略.
針對(duì)上述定理的應(yīng)用分別給出其數(shù)值例子.
例1
因?yàn)?A+BCAD)BCAπ=O,(BCAπ)2=O,
所以滿足定理1的條件,
從而,由定理1得:
因?yàn)锳AπBC=O且CAπBC=O.
所以滿足定理2的條件,從而,由定理2得
因?yàn)?CAπBCAπ=O(A+BCAD)BCAπ=O,所以滿足定理3的條件,從而,由定理3得:
因?yàn)?A+ADBC)AπBC=O 且(AπBC)3=O,所以滿足定理4的條件.從而,由定理4得
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