Unformatted text preview: Notes on linear algebra (Monday 17th October. 2016. 23:10) page 26 Proof of Proposition 2.29. By the definition of )3, we have kzbak: bap+bap+1+u s+baq=b(ap+ap+1+‘~+aq):bzak. ‘i
(by the deﬁnition of E a k)
k=a ap, ap+1, . . ., aq be some numbers. Let hp. bp+1w . . , bq be some numbers. Then, ‘5'
Z ()=ﬂk+bk kZHk-I-kZbk. k=p Proposition 2.30. Let p and a be two integers such that p < q —|— 1. Let ‘i ‘i ii ‘i
(The expression 2 ak + E bk has to be read as ( E ak) + ( E by.) .)
ic=p k k=p k=p =F’ Proof of Proposition 2.30. By the definition of Z, we have ...
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