Proving Isomorphisms of Vector Spaces

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The discussion focuses on proving that the composition of two isomorphisms, G and F, results in another isomorphism from vector space X to vector space Z. It establishes that both G and F are bijective, thus ensuring that G∘F is surjective and injective. The proof requires demonstrating that G∘F preserves vector operations, specifically through the linearity condition G∘F(au + bv) = aG∘F(u) + bG∘F(v). The conversation emphasizes the importance of understanding the properties of isomorphisms in the context of vector spaces.

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Homework Statement



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The Attempt at a Solution



isomoprism is bijective

i have no clue whatsoever..im going to research a little bit now
ill appreciate any help
 
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You are given that F is an isomorphism from X to Y and G is an isomorphism from Y->Z. You are asked to show that G[itex]\circ[/itex]F is an isomorphism from X to Z. I can guess that X, Y, and Z are vector spaces since another part of the problem talks about a "spanning set" but you should have said that! Proving an isomorphism of groups, rings, fields, etc. is quite different. An isomorphis is, as you say, bijective: you must prove this is surjective: that if z is any member of Z, then there exist x in X such that G[itex]\circ[/itex]F(x)= z. Since G is an isomorphism from Y to Z it is surjective: what does that tell you? Once you have that, you know that F is surjective from X to Y. Use that.

An isomorphism must be injective also. If x1 and x2 are such that G[itex]\circ[/itex]F(x1)= G[itex]\circ[/itex](x2) then you must prove that x1= x2. Use can use the fact that F and G are each injective to prove that. Of course, a isomorphism must also "preserve" the operations. You need to show that G[itex]\circ[/itex]F(au+ bv)= aG[itex]\circ[/itex]F(u)+ bG[itex]\circ[/itex]F(v).

As for b) using the definition of "span" together with that last statement: G[itex]\circ[/itex]F(au+ bv)= aG[itex]\circ[/itex]F(u)+ bG[itex]\circ[/itex]F(v) should be enough.
 

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