prove that the upto isomorphism their is only one vector space of dimension n.​

1 thought on “prove that the upto isomorphism their is only one vector space of dimension n.​”

1. Step-by-step explanation:

   Define a map T:V→Rn by sending each vector v∈V to its coordinate vector [v]B with respect to the basis B.

   More explicitly, if

   v=c1v1+⋯+cnvn with c1,…,cn∈R,

   then the coordinate vector with respect to B is

   [v]B=[ c1 c2 ⋮ cn ]∈Rn.

   Then the map T:V→Rn is defined by

   T(v)=[ c1 c2 ⋮ cn ].

   It follows from the properties of the coordinate vectors that the map T is a linear transformation.

   We show that T is bijective, hence an isomorphism.

   T is injective.

   To show that T is injective, it suffices to show that the null space of T is trivial: N(T)={0}.

   (See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for a proof of this fact.)

   If v∈N(T), then we have

   0=T(v)=[v]B.

   So the coordinate vector of v is zero, hence we have

   v=0v1+⋯+0vn=0.

   Thus, N(T)={0}, and T is injective.

   T is surjective.

   To show that T is surjective, let

   a=[ a1 a2 ⋮ an ]

   be an arbitrary vector in Rn.

   Then consider the vector

   v:=a1v1+⋯+anvn

   in V.

   Then it follows from the definition of the linear transformation T that

   T(v)=[v]B=[ a1 a2 ⋮ an ]=a.

   Therefore T is surjective.

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