We define a notion of isomorphism of vector spaces. Isomorphic vector spaces will have the ‘same’ vector space structures.
Definition5.1.1.
Let \(T\colon V\to W\) be a linear transformation of vector spaces over a field \(F\text{.}\) We say that \(T\) is an isomorphism over \(F\) (resp., \(F\)-isomorphism) between \(V\) and \(W\) if \(T\) is bijective.
Vector spaces \(V\) and \(W\) are said to be isomorphic if there is a bijective \(F\)-linear transformation from \(V\) onto \(W\text{.}\)
Isomorphic vector spaces \(V,W\) are denoted by \(V\simeq W\text{.}\) The set of all isomorphisms between \(V\) and \(W\) is denoted by \(\Iso_F(V,W)\text{.}\) We denote \(\Iso_F(V,V)\) by \(\Aut_F(V)\) and it is called \(F\)-automorphisms of \(V\text{.}\)
Lemma5.1.2.
Let \(T\colon V\to W\) be a bijective \(F\)-linear transformation. The set-theoretic inverse of \(T\) is also an \(F\)-linear transformation.
Let \(S\colon W\to V\) be the set-theoretic inverse of \(T\text{.}\) We need to show that for any \(\alpha,\beta\in W\) and any \(x,y\in W\text{,}\)\(S(\alpha x+\beta y)=\alpha S(x)+\beta S(y)\text{.}\)
Since \(T\) is bijective, there exists unique \(a,b\in V\) such that \(T(a)=x\) and \(T(b)=y\text{.}\) Further, the \(F\)-linearity and bijectivity of \(T\) implies that \(T(\alpha a)=\alpha T(a)=\alpha x\) and \(T(\beta b)=\beta T(b)=\beta y\text{.}\) Thus, we also get \(T(\alpha a+\beta b)=\alpha x+\beta y\text{.}\) Hence,
In view of Lemma 5.1.2, an \(F\)-linear map \(T\colon V\to W\) is an \(F\)-isomorphism if and only if there exists an \(F\)-linear map \(S\colon W\to V\) such that \(T\circ S=1_W\) and \(S\circ T=1_V\text{.}\)
Lemma5.1.4.
If \(T\colon V\to W\) is an injective \(F\)-linear transformation then, \(V\simeq\Im(T)\text{.}\)
Let \(V, W\) be finite-dimensional vector spaces over a field \(F\) of the same dimension, and let \(T\colon V\to W\) be an \(F\)-linear map. If \(T\) is injective (resp., surjective) then \(T\) is an \(F\)-isomorphism.
Let \(T\colon V\to W\) be an \(F\)-isomorphism. If \(\{v_1,v_2,\ldots,v_n\}\) is a basis of \(V\) then \(\{T(v_1),T(v_2),\ldots,T(v_n)\}\) is a basis of \(W\text{.}\) In particular, isomorphic vector spaces have the same dimension.
Let \(V=F^n\) and \(W=M_{n\times 1}(F)\) be vector spaces over a field \(F\text{.}\) Let \(e_i=(\delta_{i1},\delta_{i2},\ldots,\delta_{in})\text{.}\) Consider ordered bases \(\mathfrak{B}_V=\{e_i:1\leq i\leq n\}\) and \(\mathfrak{B}_W=(e_i^t:1\leq i\leq n)\) of \(V\) and \(W\text{,}\) respectively. Then, linear transformation \(T\colon V\to W\) given by \(x\mapsto x^t\) is an \(F\)-isomorphism.
Proposition5.1.9.
Every finite-dimensional vector space over a field \(F\) of dimension \(n\) is isomorphic to \(F^n\text{.}\)
Let \(V\) be a vector space over \(F\text{,}\) and let \(\dim_FV=n <\infty\text{.}\) Fix an ordered basis \((v_1,\ldots,v_n)\) for \(V\) and consider the standard ordered basis \((e_1,e_2,\ldots,e_n)\) for \(F^n\) (refer Example 3.6.1). By Exercise 3.5.11, for every \(v\in V\) there exists unique scalars \(\alpha_i\) such that \(v=\sum_i\alpha_iv_i\text{.}\) Thus, the map
is well-defined. It is also easy to verify that \(T\) is an \(F\)-isomorphism.
Lemma5.1.10.
Let \(\{v_1,v_2,\ldots,v_n\}\) and \(\{w_1,w_2,\ldots,w_n\}\) be bases of \(V\) and \(W\text{,}\) respectively. Consider the mapping \(T\colon V\to W\) such that \(v_i\mapsto w_i\text{.}\) Extend the map linearly on \(V\) (see Remark 4.1.5). Then, the map \(T\) is an isomorphism.
Remark5.1.11.
We observed earlier (Remark 4.1.8) that \(\End_F(V)\) is a ring. We denote the group of all units in \(\End_F(V)\) by \(\GL(V)\text{.}\) The group \(\GL(V)\) is called the general linear group of \(V\text{.}\) So, an element of \(\GL(V)\) is an \(F\)-isomorphism of \(V\) onto itself.
