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