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Section 5.1 Definition of Isomorphism

We define a notion of isomorphism of vector spaces. Isomorphic vector spaces will have the ‘same’ vector space structures.
Definition 5.1.1.

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 \(V\text{.}\)

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,

\begin{align*} S(\alpha x+\beta y)\amp=S\big(T(\alpha a+\beta b)\big)\\ \amp=\alpha a +\beta b\\ \amp=\alpha S\big(T(a)\big)+\beta S\big(T(b)\big)\\ \amp=\alpha S(x)+\beta S(y). \end{align*}
Remark 5.1.3.
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 \(S\circ T=1_V\text{.}\)
A linear functional \(f\colon V\to F\) (see Definition 4.1.6) is an isomorphism if and only if \(V\simeq F\text{.}\)
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.
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
\begin{equation*} T\colon V\to F^n\quad\text{given by}\quad \sum\alpha_iv_i\mapsto \sum\alpha_ie_i \end{equation*}
is well-defined. It is also easy to verify that \(T\) is an \(F\)-isomorphism.
Remark 5.1.11.
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.