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What else can I help you with?
Is etanercept or enbrel made from fetal lung cells referred to by the code WI 26 VA4?
Use of Human Cell Lines in PharmCheck it out. It seems it does come from the aborted fetuses. These multinational corporations certainly seem to worship Lucifer.
How do you show the direct sum of the image and kernel of the linear operator is the vector space?
Ok, linear algebra isn't my strongest area but I'll have a go (please note that all vectors are in bold, if a mathematical lower case letter is not in bold, assume it to be a scalar unless otherwise indicated).First, I'm going to assume that you already know how to prove that the kernel and image of a linear operator, say f:U->V where U and V are vector spaces, is a vector subspace of U as this needs to be proved before going on to show that the sum of ker(f) and Im(f) are a vector space.Now, a vector space must follow the following axioms (here A represents the addition axioms, and M the multiplication axioms):A1) associativity u+(v+w)=(u+v)+wA2) commutativity u+v=v+uA3) identity there exists an element 0 in U such that v+0=vA4) existence of an inverse for all u in U there exists an element -u in U such that u+(-u)=0M1) scalar multiplication with respect to vector addition a(u+v)=au+avM2) scalar multiplication with respect to field addition (a+b)u=au+buM3) compatibility of scalar and field multiplication (ab)u=a(bu)M4) identity 1u=u where 1 is the multiplicative identityker(f)={u in U | f(u)=0} (or more strictly, the set of vectors u in U such that Au=0 i.e. ker(f) is the null space of the matrix A), Im(f)={v in V | f(u)=v for some u in U}Let ker(f)+Im(f) = W. Show W is a vector space i.e. show W satisfies the above axioms.A1, A2, M1, M2, and M3 are all trivial/simple to prove.A3: since Im(f) is the set off all v in V obtained from f(u), assume Im(f) = V and as V is a vector space it must have, by definition, an additive identity 0, therefore W also contains 0.A4: since ker(f) is the null space of the matrix A, f(0)=0. Now assume that f(u)=0 and f(-u)=v, since U is a vector space u+(-u)=0 so f(u+(-u))=f(0)=0 but f(u+(-u))=f(u)+f(-u)=0+vtherefore, 0=0+v which implies v=0, so if u goes to 0 in V, its inverse -u also goes to 0. A similar process for the image shows that if u goes to v then -u goes to -v. (Thanks to Jokes Free4Me for the help with this axiom)M4: similar to A3 except V, again by definition, must have the multiplicative identity 1 and so also exists in W.All the axioms are satisfied, therefore W=ker(f)+Im(f) is a vector space. Q.E.D.
