Question

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

Answer 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.