Invertible linear maps, isophormism and isomophic. I am reading Sheldon Axler's book on linear algebra. However, the concept of isomorphic is not clear to me.
The book says: "Two vector spaces are called isomorphic if there is an isomorphism
from one vector space onto the other one".
My question is: if two vector spaces are isomorphic, are there two invertible linear maps that link both vector spaces? For example, if V and W are two isomorphic vector spaces, then there exists:
$$T: V -> W$$ and $$ S: W->V$$ such that:
$$ST=I$$ and $$TS=I$$
Is correct?
Another question, if I want to prove that two vector spaces are isomorphic, is it enough to prove that one of the linear maps that relate them is bijective?
 A: That's the point. An isomorphism between two finite dimensional vector spaces is a bijective linear transformation $T:V\to W$. It can be proven that $T^{-1}$ is also a linear mapping.
Indeed, let $w_{1}\in W$, $w_{2}\in W$ and $a\in\textbf{F}$.
Since $T$ is surjective, there are $v_{1}\in V$ and $v_{2}\in V$ such that $T(v_{1}) = w_{1}$ and $T(v_{2}) = w_{2}$.
Moreover, since $T$ is injective, such vectors $v_{1}$ and $v_{2}$ are unique.
Consequently, once $T(av_{1} + v_{2}) = aw_{1} + w_{2}$, we conclude that
\begin{align*}
T^{-1}(aw_{1} + w_{2}) = av_{1} + v_{2} + aT^{-1}(w_{1}) + T^{-1}(w_{2})
\end{align*}
and $T^{-1}$ is linear, just as desired.
Another way to prove that two finite dimensional vector spaces are isomorphic consists in proving they have the same dimension.
Let us prove that $\dim V = \dim W$ implies that $V$ and $W$ are isomorphic.
Since $\dim V = \dim W < \infty$, we can assume that $\dim V = \dim W = n$.
Thus can choose two basis $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$ and $\mathcal{B}_{W} = \{w_{1},w_{2}\ldots,w_{n}\}$
Then we can define a linear transformation between $V$ and $W$ according to the realtion $T(v_{j}) = w_{j}$.
In this case, $T$ is injective and, due to the rank-nullity theorem, it is a bijection as well.
Then we shall consider 
\begin{align*}
& a = a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n}\in V\\\\
& b = b_{1}v_{1} + b_{2}v_{2} + \ldots + b_{n}v_{n}\in V
\end{align*}
Hence we get
\begin{align*}
T(a) = T(b) & \Rightarrow T(a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n}) = T(b_{1}v_{1} + b_{2}v_{2} + \ldots + b_{n}v_{n})\\\\
& \Rightarrow a_{1}T(v_{1}) + a_{2}T(v_{2}) + \ldots + a_{n}T(v_{n}) = b_{1}T(v_{1}) + b_{2}T(v_{2}) + \ldots + b_{n}T(v_{n})\\\\
& \Rightarrow a_{1}w_{1} + a_{2}w_{2} + \ldots + a_{n}w_{n} = b_{1}w_{1} + b_{2}w_{2} + \ldots + b_{n}w_{n}\\\\
& \Rightarrow (a_{1} - b_{1})w_{1} + (a_{2} - b_{2})w_{2} + \ldots + (a_{n} - b_{n})w_{n} = 0 \Rightarrow a = b
\end{align*}
Conversely, if $V$ and $W$ are isomorphic, then $\dim V = \dim W < \infty$
The results which allow us to say so are described next
Lemma 1
Suppose that $S = \{s_{1},s_{2},\ldots,s_{m}\}$ spans the vector space $V$. Then any set of linear independent vectors has no more than $m$ vectors.
Lemma 2
If the linear transformation $T:V\to W$ is injective, then it takes LI vectors onto LI vectors.
Proposition
If there is an isomorphism between the finite dimensional vector spaces $V$ and $W$, then $\dim V = \dim W$.
Proof
Let us consider a basis $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{m}\}$ and a basis $\mathcal{B}_{W} = \{w_{1},w_{2},\ldots,w_{n}\}$.
Once $T$ is injective, we conclude that $T(\mathcal{B}_{V}) = \{T(v_{1}),T(v_{2}),\ldots,T(v_{m})\}$ is LI. Similarly, once $T^{-1}$ is injective, we conclude that $T(\mathcal{B}_{W}) = \{T^{-1}(w_{1}),T^{-}(w_{2}),\ldots,T^{-1}(w_{n})\}$ is LI. But $\mathcal{B}_{W}$ spans $W$. Thus $m\leq n$. Correspondingly, $\mathcal{B}_{V}$ spans $V$. Thus $n\leq m$. Thus $n = m$ and we are done.
A: Yes, that's right. For the sake of completeness, here I repeat the definition:

Let $V,W$ be vector spaces over a field $\Bbb{F}$. We say $V$ is isomorphic to $W$ (as a vector space) if there is an invertible function $T:V \to W$ (with inverse denoted as $T^{-1}:W \to V$) such that both $T$ and $T^{-1}$ are linear.
In this case, we call $T$ an isomorphism (from $V$ onto $W$).

In the above definition, I put "(as a vector space)" in brackets, because the word "isomorphism" is also used in other contexts (for example, group isomorphims, field isomorphisms, isomorphism of inner-product spaces etc), and they have slightly different meaning. But as long as you're in the realm of linear algebra and vector spaces, you can safely omit that phrase, and there will be no misunderstanding.
One can easily check that if $T:V \to W$ is linear and invertible, then the inverse $T^{-1}:W \to V$ is automatically linear. Once you prove this (simple) theorem, as long as you check that $T:V \to W$ is linear and bijective (i.e invertible), you never again have to check whether or not the inverse is also linear, because it will be.
Another thing which is easy to prove is that "is isomorphic to" is an equivalence relation:

*

*$V$ is isomorphic to $V$.

*If $V$ is isomorphic to $W$ then $W$ is isomorphic to $V$.

*If $V$ is isomorphic to $W$ and $W$ is isomorphic to $X$ then $V$ is isomorphic to $X$.

As a result of this, rather than saying "$V$ is isomorphic to $W$", we sometimes use the terminology "$V$ and $W$ are isomorphic" or "$W$ and $V$ are isomorphic". The symmetry of the relation means we don't have to keep track of the order in which we say things.
