2
$\begingroup$

I am really confused by this concept of isomorphism, it seems to be a new name for something that is already well understood. Every time I look up the definition for isomorphism, the definition changes. In other fields, they are called homomorphism, which confuses with homeomorphism (let's not get into that). They are of course equivalent, but the argument is quite subtle.

Acccording the following reliable source (Naylor and Sell):

Given linear spaces $X, Y$ over $ \mathbb{F}$$, T: X \to Y$ is an isomorphism iff (i) $T$ is $1-1$, (ii) $T$ is onto, (iii) $T$ is linear, (iv) $T^{-1}$ is linear

Ok, nice and simple! Computationally tractable. Something you would use on your exams to check if $T$ is an isomorphism.

But when you look at the other discussions on Math.SE, there seems to be a multi-panel debate as to what an isomorphism even is.

For example: What's the difference between a bijection and an isomorphism?

First answer: "As another example, if the sets are vector spaces, then an isomorphism is a bijection that preserves vector addition and scalar multiplication."

Or here: What's the difference between a bijection and an isomorphism?

"The answer is "vice versa." An isomorphism is a structure-preserving bijection. The specific meaning of "structure" will vary, depending on the context."

Where are people getting their definition from? At no point in definition given above was anything said about "preservation of vector addition and scalar multiplication". Why is that so?

$\endgroup$
  • 2
    $\begingroup$ "(iii) $T$ is linear" is definitely a statement about preserving vector addition and scalar multiplication. $\endgroup$ – Hoot Oct 25 '15 at 2:10
  • $\begingroup$ @Hoot Yes but linearity is a property of the transformation, how does it relate back to $+$, and $\times$ which we equip with the underlying set which then forms the vector space? $\endgroup$ – Carlos - the Mongoose - Danger Oct 25 '15 at 2:13
  • 1
    $\begingroup$ Linearity isn't just a property of the transformation; it's a property of the transformation and the operations on each vector space. If you pretend the transformation is just the identity map, then linearity is just the statement that the operations on the two vector spaces are the same. $\endgroup$ – Eric Wofsey Oct 25 '15 at 2:20
  • 1
    $\begingroup$ @FenceJumper: There are some good reasons to say "isomorphism", but ultimately, it is just a convention. Why do we say "orange" instead of "reddish yellow"? It's just the word for it. You should not always expect language (which is ultimately just a social convention) to be perfectly logical. $\endgroup$ – Eric Wofsey Oct 25 '15 at 2:44
  • 2
    $\begingroup$ In linear algebra an isomorphism is simply a linear bijection between vector spaces. One source of confusion in math is that the same word is sometimes used to mean different things, and this is the case with the word "isomorphism" which has different meanings for different types of algebraic structures (such as groups, rings, vector spaces...) Perhaps it would be more clear to call it a "vector space isomorphism" or a "group isomorphism" or a "ring isomorphism", never just "isomorphism". In each case we have a bijection that "preserves" relevant structure. $\endgroup$ – littleO Oct 25 '15 at 3:06
2
$\begingroup$

What does it mean for $T$ to be linear?

It precisely means given $u$ and $v$ in $X$ and $\alpha \in \Bbb F$,

  1. $T(u + v) = T(u) + T(v)$ (i.e., $T$ preserves vector addiction).
  2. $T(\alpha u) = \alpha T(u)$ (i.e., $T$ preserves scalar multiplication).

So the definition you listed does state that $T$ must preserve vector addition and scalar multiplication, but it does so concisely by saying "$T$ is linear".

$\endgroup$
  • 1
    $\begingroup$ Is "preserve" a common technical terminology or is it used informally? I guess part of my confusion is related to not understanding what it is meant for some operation to preserve another operation $\endgroup$ – Carlos - the Mongoose - Danger Oct 25 '15 at 2:33
  • $\begingroup$ @FenceJumper In this context, the word "preserve" means this: we know $X$ has an addition operation $+_{X}$, and $Y$ has an addition operation $+_{Y}$. We say a map "preserves" addition between $X$ and $Y$ if the $+_{X}$ sum of two elements in $X$ always maps to the $+_{Y}$ sum of the images of the elements, i.e., $T(u +_{X} v) = T(u) +_{Y} T(v)$. In other words, we are assured if $u$ and $v$ act in a certain way under $+_{X}$, then $T(u)$ and $T(v)$ will act in a similar way under $+_{Y}$, and by similar way, I mean exactly that $T(u) +_{Y} T(v)$ will behave exactly as $T(u +_{X} v)$. $\endgroup$ – layman Oct 25 '15 at 2:38
  • $\begingroup$ @FenceJumper Think of e.g. "preserves addition" as shorthand for "preserves equations involving addition" - if $x+y=z$ is true in $X$, then $T(x)+T(y)=T(z)$ is true in $Y$. And I would say this is a piece of common mostly technical terminology. $\endgroup$ – Noah Schweber Oct 29 '15 at 5:06
2
$\begingroup$

A vector space is an additive commutative group $V$ along with an action of a field $k$ on $V$: for each scalar $\lambda\in k$ there is a map $\mu_\lambda : V\mapsto V$ that sends $v\mapsto \lambda v$. Since $V$ is also a commutative group under addition, we have a binary operation $\sigma: V\times V\to V$ that sends a pair $(v,w)\mapsto v+w$. Note then that a linear transformation between a space $V$ and a space $W$ is a map that commutes with this collection of "structure" functions on $V$ and $W$. We want that, for each $\lambda\in k$, $$f(\lambda v)=f(\mu_\lambda(v))=\mu'_\lambda(f(v))=\lambda f(v)$$

$$f(v+w)=f(\sigma(v,w))=\sigma'(f(v),f(w))=f(v)+f(w)$$

where $\mu$ are the multiplication by scalars in $V$ and $\sigma$ the sum of $V$, and $\mu',\sigma'$ the corresponding operations in $W$. It is in this sense that we say $f$ preserves the structure of vector spaces.

One usually has a collection of objects, in your case vector spaces, and a collection of arrows from and to objects. An isomorphism is defined to be an arrow $f$ that admits an inverse that is also an arrow. In your case, one can check that a linear transformation admitting an inverse is the same as the linear transformation itself being a bijection as a map of sets, and hence the result. In general, there are collection of objects, such as topological spaces, where there are arrows with no inverse arrow but such that, as a function, this arrow is bijective. A classical example is the map that wraps the unit interval $[0,1)$ onto the unit circle $S^1$, by sending $t\mapsto e^{2i t\pi}$. This is continuous, injective and onto, but its inverse is not continuous, and hence is not a map in the category (I could avoid the word for so long) of topological spaces.

$\endgroup$
0
$\begingroup$

The proper and universal definition of an isomorphism (in any given category) is that it is a morphism between objects that admits an inverse morphism. (If such an inverse exists it will always be unique.) The category will tell you what are (homo)morphisms; in the category of $F$-vector spaces they are $F$-linear maps. That morphisms preserve structure is just an informal statement, which really means that we call "structure" that what is preserved by morphisms. For instance instance in vector spaces having a linear dependency between some vectors implies having the same relation between their image across a linear map, so such relations are part of the structure in vector spaces.

Now in many categories morphisms are maps, and maps having inverses must be bijections. And often the conditions of being a morphism and bijective implies that the inverse map is also a morphism; this is true in the category of $F$-vector spaces (which is a nice exercise to prove), and in many other algebraic categories (but not for instance in the category of topological spaces where morphisms are continuous maps). When it holds one will often find instead of the above "high definition" of isomorphism the "low definition" of an isomorphism as a bijective morphism. It can be of some advantage in practice to not have to check explicitly that the inverse map of a morphism is a morphism, but still I feel that this should be formulated as a proposition, not integrated into the definition of isomorphism. I suppose it is done this way because students have a strong tendency to use the definition instead of trying to apply a proposition.

Your book more or less follows high definition, but instead of just asking for the existence of an inverse linear map (which would imply (i) and (ii), and also obviously has to coincide with $T^{-1}$) prefers to spell out conditions that imply the existence of an inverse map $T^{-1}$ (this is the only reason that (i) and (ii) are needed) and then require that inverse map to be linear (iv).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.