Hiu Nga--Cheung>>hi\ TA Bong_Lee>>hi TA Bong_Lee>>your mid term ok? Hiu Nga--Cheung>>i don't know Hiu Nga--Cheung>>i post a Q in discussion board TA Bong_Lee>>let me see TA Bong_Lee>>i guess the question is wrong where is the prob come from? Hiu Nga--Cheung>>Rank and Nullity Hiu Nga--Cheung>>2. Properties of rank and nullity Hiu Nga--Cheung>>bottom TA Bong_Lee>>i think it just a typing mistake change m to n! Hiu Nga--Cheung>>Dual basis Hiu Nga--Cheung>>i don't understand the whole page Hiu Nga--Cheung>>what is isomorphism TA Bong_Lee>>just a moment TA Bong_Lee>>ok TA Bong_Lee>>the definition of isomorphism between two set is the bijection mapping between two set such that operation preserve TA Bong_Lee>>let me make an example TA Bong_Lee>>let A be the subset of 2by 2 matrix in this form TA Bong_Lee>>[a 0] TA Bong_Lee>>[0 0] TA Bong_Lee>>let R be the set of real no. TA Bong_Lee>>f:R ->A f(a)=[a 0] TA Bong_Lee>>[0 0] TA Bong_Lee>>so that we have TA Bong_Lee>>f(a+b)=f(a)+f(b) TA Bong_Lee>>so addition is preserved TA Bong_Lee>>and f is clearly a bijection TA Bong_Lee>>so f is isomorphism under addition TA Bong_Lee>>infact the true definition is define on group not on set but just give u some feeling Hiu Nga--Cheung>>f Õ (f(b1), f(b2), ..., f(bn)), V* Õ Rn Hiu Nga--Cheung>>in the blue box Hiu Nga--Cheung>> what does it mean TA Bong_Lee>>ok TA Bong_Lee>>I let that function F:V*->Rn TA Bong_Lee>>by define F(f)->(f(b1),f(b2),...,f(bn)) TA Bong_Lee>>V* is set of all linear function TA Bong_Lee>>b1,b2....bn are fixed TA Bong_Lee>>so what we inout is a liner function TA Bong_Lee>>and we get back an element in Rn TA Bong_Lee>>anything uncleear? Hiu Nga--Cheung>>the proof below Hiu Nga--Cheung>>fa(c1b1 + c2b2 + ... + cnbn) = a1c1 + a2c2 + ... + ancn TA Bong_Lee>>u r asking y they are equal? Hiu Nga--Cheung>>what is fa Hiu Nga--Cheung>>is V: c1b1 + c2b2 + ... + cnbn Hiu Nga--Cheung>>and why they are equal TA Bong_Lee>>First we r now proofing that E is bijective TA Bong_Lee>>and the part u mentioned is for proofing E is surjective Hiu Nga--Cheung>>surjective??? TA Bong_Lee>>yes Hiu Nga--Cheung>>what is it TA Bong_Lee>>E is a mapping from set of linear function to Rn TA Bong_Lee>> a mapping say f;X->Y is surjective means for every element y in Y there is x in X Hiu Nga--Cheung>>what is fa TA Bong_Lee>>such that f(x)=y Hiu Nga--Cheung>>why they are equal TA Bong_Lee>>ic TA Bong_Lee>>now let me explain TA Bong_Lee>>in order to prove E is surjective TA Bong_Lee>>that means if we pick up an element in Rn say (a1,a2,...an) TA Bong_Lee>>we want to find a liner transformation correspond to (a1,a2,...,an) TA Bong_Lee>>but remember in the beginning of part1 of dual space TA Bong_Lee>>the linear function on euclidean space is in the form f(a1,a 2, ...,an ) = a1c1 + a2c2 + ... + ancn TA Bong_Lee>>so we need to find fa(c1b1 + c2b2 + ... + cnbn) = a1c1 + a2c2 + ... + ancn. TA Bong_Lee>>fa is fa(bi)=ai Hiu Nga--Cheung>>why we write c1b1 + c2b2 + ... + cnbn TA Bong_Lee>>i don't know TA Bong_Lee>>maybe put it on the discussion board TA Bong_Lee>>bye