What structure is preserved in a vector space isomorphism?

Question

The definition of an isomorphism is "a structure-preserving mapping between two structures of the same type that can be reversed by an inverse mapping".

In my linear algebra class, an isomorphism is defined as "a bijective linear transformation." I understand some of the structure that is preserved between the spaces, but is there some exact fixed set of things that are preserved by a bijective linear transformation? (or an isomorphism in general). Further, before we can say that a bijective linear transformation is an isomorphism, do we not need to prove that all of the structure is actually preserved? I ask this because we seem to just assume that a lot of things are preserved between different vector spaces, but never have we proven that these things are preserved between spaces. Below are a few things that I am a little confused about is they need to be proven or follow from the fact that $T$ is a bijective linear transformation.

Let $T: V \to W$ be an isomorphism. I understand that since $T$ is linear, scalar multiplication and addition will be preserved. Must it be shown that if there is a linear transformation $A$ in $V$ such that for all $x \in V$, there exists a unique transformation $B$ in $W$ such that $A(x) = T^{-1}(B(T(x))$, or that if $S$ is a basis of $W$ of where $S = (S_1, \dots, S_n)$, then $(T^{-1}(S_1), \dots, T^{-1}(S_n))$ is a basis of $V$?

I'm not exactly sure where this preservation of structure stops, and when I am no longer using the fact that we have an isomorphism, and rather sort of just saying things that are not proven true. Maybe I am not entirely sure what constitutes as structure that needs to be preserved in terms of vector spaces.

Any help clearing up this idea of isomorphism is greatly appreciated.

Answer 1

The structure that is preserved in a vector space isomorphism are exactly the linear relations: $$ \lambda_1 v_1 + \lambda_2 v_2 + \cdots + \lambda_m v_m = 0 \text{ in } V \iff \lambda_1 w_1 + \lambda_2 w_2 + \cdots + \lambda_m w_m = 0 \text{ in } W $$ An isomorphism $V \to W$ just renames the elements of $V$ into elements of $W$ subject to the condition above.

Make sure you understand that the claim above is equivalent to the standard definition: $T: V \to W$ is an isomorphism iff $T$ is a bijection such that $T(\lambda v)=\lambda T(v)$ and $T(v_1 + v_2) = T(v_1) + T(v_2)$.

Answer 2

Set isomorphisms are just bijective functions and you can think of them as a relabeling of their elements. Algebraic isomorphisms are a relabeling that also map the binary operations in a way that allows you to do the algebra in either set and translate back and forth between them. For vector spaces this will look like $f(x+y)=f(x)+f(y)$ and $f(cx)=cf(x)$ where the addition and multiplication on both sides of the equations are each done in their own space. They can be entirely different in their computational properties but the isomorphic mapping keeps them compatible with each other. Linear transformations in general are homomorphisms and bijective linear transformations are isomorphism. They just use different terminology for historical reasons just as the nullspace is a linear version of the kernel of a homomorphism.

Answer 3

In general, what you really have is some many-sorted FOL (first-order logic) structure. For example, a vector space is nothing more than a two-sorted FOL structure, with one sort $V$ of vectors and the other sort $F$ of scalars, and with four operations $⊕,•,+,·$, and typically also with three constants $z,0,1$ such that:

1. $⊕ : V^2{→}V$.
2. $• : F{×}V{→}V$.
3. $(V,⊕,z)$ is a commutative group.
4. $(F,+,·,0,1)$ is a field.
5. $1•x = x$ for every $x{∈}V$.
6. $(a·b)•x = a•(b•x)$ for every $a,b{∈}F$ and $x{∈}V$.
7. $(a+b)•x = (a•x)⊕(b•x)$ for every $a,b{∈}F$ and $x{∈}V$.
8. $a•(x⊕y) = (a•x)⊕(a•y)$ for every $a{∈}F$ and $x,y{∈}V$.

An isomorphism between (many-sorted) FOL structures is just a bijective mapping of all the elements of each sort in one structure to the elements of the corresponding sort in the other structure such that the mapping commutes with the operations. As an example, an isomorphism between vector spaces $S = (V,F,⊕,•,+,·)$ and $S' = (V',F',⊕',•',+',·')$ is a pair $(i,j)$ such that:

1. $i$ bijects from $V$ to $V'$.
2. $j$ bijects from $F$ to $F'$.
3. $i(x⊕y) = i(x)⊕'i(y)$ for every $x,y{∈}V$.
4. $i(a•x) = j(a)•'i(x)$ for every $a{∈}F$ and $x{∈}V$.
5. $j(a+b) = j(a)+'j(b)$ for every $a,b{∈}F$.
6. $j(a·b) = j(a)·'j(b)$ for every $a,b{∈}F$.

That's precisely all there is to isomorphisms in general. Note that for vector spaces in particular, we typically focus on vector spaces over a fixed field $F$, and $F$ is often either $ℝ$ or $ℂ$, in which case we have strictly more than a field structure. In particular, $(ℝ,0,1,+,·)$ is a complete ordered field, and any isomorphism from a complete ordered field to itself is the identity, and $ℂ$ is built from $ℝ$, so when we talk about isomorphism between vector spaces $S$ and $S'$ where $F$ is either $ℝ$ or $ℂ$, we want the bijection $j$ to actually be the identity, and that is why we are left with just the criterion that $i$ commutes with the operations involving vectors:

1. $i(x⊕y) = i(x)⊕'i(y)$ for every $x,y{∈}V$.
2. $i(a•x) = a•'i(x)$ for every $a{∈}F$ and $x{∈}V$.

This in turn is equivalent to the isomorphism being a bijective linear transformation!

By the way, if you apply this general viewpoint based on many-sorted FOL structure axiomatization to other algebraic concepts as well, you will see that many concepts that appear different for different kinds of structures are actually identical once expressed from the logical viewpoint. For example, see this post about quotient structures and how to derive the criteria for 'quotienting'.

Answer 4

(This is a response to a question OP asks in the comments which is too long to keep in the comments, but is still relevant as an answer to the original question.)

Model Theory and/or Universal Algebra provide a very general way of approaching the question you're asking about what 'structure' is and what it means for 'isomorphisms' to preserve 'properties' of that structure. The other questions have provided more concrete explanations of this for the specific case of vector spaces, so I'll give a more abstract answer. I will also be approaching $\mathbb{F}$-vector spaces as a single-sorted structure, in contrast to user21820's approach.

A (first-order) language is a triple $\langle \mathcal{F}, \mathcal{R}, \operatorname{ar}\rangle$ consisting of disjoint sets $\mathcal{F}$ and $\mathcal{R}$ and a function $\operatorname{ar} \colon \mathcal{F} \sqcup \mathcal{R} \to \mathbb{N}$. Our interpretation of this triple is that $\mathcal{F}$ denotes a set of function symbols, $\mathcal{R}$ denotes a set of relation symbols, and $\operatorname{ar}$ associates to each function and relation symbol an arity.

In our particular example of $\mathbb{F}$-vector spaces (where $\mathbb{F}$ is any field, e.g., $\mathbb{R}$ or $\mathbb{C}$), we can take $\mathcal{F} = \{+\} \cup \{s_a \mid a \in \mathbb{F}\}$, $\mathcal{R} = \emptyset$, and $\operatorname{ar}(+) = 2$ and $\operatorname{ar}(s_\alpha) = 1$ for each $\alpha \in \mathbb{F}$. $+$ has its usual interpretation as vector addition, while the function symbols $s_a$ correspond to scalar multiplication by $a$ (this is how we get around the 'two-sorted' nature of vector spaces after fixing a field). Optionally you can include a $0$-ary function (i.e., constant) symbol $0$ for the zero vector and/or a unary function symbol $-$ for vector negation, but these are 'definable' in the sense I'll indicate below.

Given a language $\langle \mathcal{F}, \mathcal{R}, \operatorname{ar}\rangle$, a structure over that language is a triple $\mathbf{A} = \langle A, \langle f^\mathbf{A}\rangle_{f \in \mathcal{F}}, \langle R^\mathbf{A} \rangle_{R \in \mathcal{R}}\rangle$ such that for each $f \in \mathcal{F}$, $f^\mathbf{A}$ is a function $A^{\operatorname{ar} f} \to A$ and for each $R \in \mathcal{R}$, $R^\mathbf{A}$ is a $(\operatorname{ar} R)$-ary relation over $A$. In other words, a structure over a language is a set and instantiations of all the function and relation symbols (such that the arities line up appropriately). Note that at this point we haven't introduced any 'axioms' or the like.

If $\mathbf{A}$ and $\mathbf{B}$ are two structures over a language $\langle \mathcal{F}, \mathcal{R}, \operatorname{ar}\rangle$, an isomorphism between $\mathbf{A}$ and $\mathbf{B}$ is a bijection $\varphi \colon A \to B$ such that for function symbol $f \in \mathcal{F}$ and relation symbol $R \in \mathcal{R}$ (say, where $\operatorname{ar} f = \operatorname{ar} R = n$) and $a_1,a_2,\ldots,a_n$, $$\varphi(f^\mathbf{A}(a_1,a_2,\ldots,a_n)) = f^\mathbf{B}(\varphi(a_1),\varphi(a_2),\ldots,\varphi(a_n))$$ and $$\langle a_1,a_2,\ldots,a_n\rangle \in R^\mathbf{A} \iff \langle \varphi(a_1),\varphi(a_2),\ldots,\varphi(a_n)\rangle \in R^\mathbf{B}.$$ In other words, an isomorphism is a bijection which is compatible with the operations (or commutes with the operations) and both preserves and reflects each relation.

In our particular example of $\mathbb{F}$-vector spaces with the language I gave above, a structure over the language of $\mathbb{F}$-vector spaces is just a set $V$, a binary operation $+\colon V^2 \to V$, and unary operations $s_\alpha \colon V \to V$. An isomorphism $\varphi$ between such structures asks $\varphi$ to be a bijection and for $\varphi(v+w) = \varphi(v)+\varphi(w)$ and $\varphi(s_\alpha(v)) = s_\alpha(\varphi(v))$ for all $v,w$ in the domain of our first structure and $\alpha \in \mathbb{F}$. So far, this doesn't look quite like the typical definition of an isomorphism of $(\mathbb{F})$-vector spaces which would ask that $\varphi$ satisfy $\varphi(s_\alpha(v)+w) = s_\alpha(\varphi(w)) + \varphi(w)$ for all $v,w \in V$ and $\alpha \in \mathbb{F}$. We'll address that now.

We like to deal with structures that exhibit particular properties, like groups, rings, posets, etc. One way to realize this in the framework above is to define a first-order theory $\Phi$ over a language $\sigma = \langle \mathcal{F},\mathcal{R},\operatorname{ar}\rangle$ to be a set of well-formed formula in that language. To give the shortest definition I can of what is meant by 'well-formed formula', fix a countably infinite set of variables $X$ and consider strings over the set $\Sigma = X \cup \mathcal{F} \cup \mathcal{R} \cup \{ \text{`('},\text{`)'},\text{`,'},=,\vee,\wedge,\to,\neg,\leftrightarrow,\forall,\exists\}$ with the usual interpretation of the logical symbols and assuming that everything is distinct. $\mathrm{Term}_\sigma$ is the smallest subset of $\Sigma^\ast$ (the set of strings built up over the alphabet $\Sigma$) such that:

1. $X \subseteq \mathrm{Term}_\sigma$,
2. if $f \in \mathcal{F}$ is $n$-ary and $t_1,t_2,\ldots,t_n \in \mathrm{Term}_\sigma$, then $f(t_1,t_2,\ldots,t_n) \in \mathrm{Term}_\sigma$ (where if $n=0$ we interpret this as just '$f$').

In other words, $\mathrm{Term}_\sigma$ is the set of expressions built up from the variables and function symbols. Then define $\mathrm{Form}_\sigma$ to be the smallest subset of $\Sigma^\ast$ such that:

1. For any terms $t,s \in \mathrm{Term}_\sigma$, $(t = s) \in \mathrm{Form}_\sigma$.
2. If $\varphi,\psi \in \mathrm{Form}_\sigma$, then $(\varphi \wedge \psi)$, $(\varphi \vee \psi)$, $(\varphi \to \psi)$, $\neg \varphi$, and $(\varphi \leftrightarrow \psi)$ are in $\mathrm{Form}_\sigma$.
3. If $\varphi \in \mathrm{Form}_\sigma$ and $x \in X$, then $\exists x \varphi$ and $\forall x \varphi$ are in $\mathrm{Form}_\sigma$.
4. If $R \in \mathcal{R}$ is $n$-ary and $t_1,t_2,\ldots,t_n \in \mathrm{Term}_\sigma$, then $R(t_1,t_2,\ldots,t_n) \in \mathrm{Form}_\sigma$.

I'll call elements of $\mathrm{Form}_\sigma$ wffs (well-formed formulas). A sentence is a wff $\varphi$ such that if $x$ is a variable appearing in $\varphi$, then it appears in a subformula of $\varphi$ of the form $\exists x \psi$ or $\forall x \psi$ -- in other words, they're wffs without free variables, or equivalently wffs all of whose variables are bound. The point of sentences is that they correspond to wffs whose 'truth' in any particular structure shouldn't depend on providing an additional assignment of values to free variables.

Finally we can say that a first-order theory is nothing more than a set $\Phi$ of sentences; in other words, we're picking out sentences which will be our 'axioms'.

In our particular example of $\mathbb{F}$-vector spaces, the theory we take consists of the following sentences/axioms (diverging a bit in notation by using infix notation):

1. $\forall v \forall w (v+w = w+v)$ (associativity of vector addition)
2. $\exists 0 \forall v ((v+0 = v) \wedge \exists w (v+w = 0))$ (existence of a zero vector and additive inverses)
3. For each $\alpha \in \mathbb{F}$, $\forall v \forall w (s_\alpha(v+w) = s_\alpha(v) + s_\alpha(w))$ (scalar multiplication distributes over vector addition)
4. For each $\alpha,\beta \in \mathbb{F}$, $\forall v ( s_\alpha(s_\beta(v)) = s_{\alpha \cdot \beta}(v))$ (compatibility of scalar multiplication with field muliplication)
5. For each $\alpha,\beta \in \mathbb{F}$, $\forall v (s_{\alpha+\beta}(v) = s_\alpha(v)+s_\beta(v))$ (compatibility of scalar multiplication with vector addition)
6. $\forall v (s_1(v) = v)$ (identity element of scalar multiplication; here $1$ is the multiplicative identity of the field)

A $\mathbb{F}$-vector space is then a structure over the aforementioned language which satisfies the above first-order theory. Giving a precise definition of satisfaction would be too long, but an intuitive idea probably suffices. It can then be easily shown that for any map $\varphi$ between vector spaces, having $\varphi(\alpha v + w) = \alpha \varphi(v)+\varphi(w)$ is equiavlent to asking for $\varphi(\alpha v) = \alpha \varphi(v)$ and $\varphi(v+w) = \varphi(v)+\varphi(w)$.

Last but not least, we double back to what all this means in terms of isomorphisms. The general result is the following:

Theorem. Given a language $\sigma = \langle \mathcal{F},\mathcal{R},\operatorname{ar}\rangle$ and two structures $\mathbf{A}$ and $\mathbf{B}$ over that language, if there exists an isomorphism from $\mathbf{A}$ to $\mathbf{B}$, then for every sentence $\varphi$ in the language $\mathbf{A}$ satisfies $\varphi$ if and only if $\mathbf{B}$ satisfies $\varphi$.

So isomorphic structures agree with each other on any property/identity/etc that can be written down as a first-order sentence.

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Disclaimer: The specific way I defined what a wff is isn't entirely universal; some approaches use Polish notation, some move $=$ from being a logical symbol to being a relation symbol and then add appropriate axioms characterizing it, some use a single variable symbol and include a way to generate infinitely many variables out of that (e.g., given $x$, we add another logical unary operation symbol $'$ such that $x'$, $x''$, $x'''$, etc. are all independent logical variables), etc. The specific ways I wrote the axioms for $\mathbb{F}$-vector spaces isn't standard (though pretty typical), nor is the language I used (as mentioned above, a constant ($0$-ary function) symbol $0$ and/or a unary funciton symbol $-$ might be used). Some folks like to distinguish constant symbols from function symbols. Etc. etc.