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I know that isomorphic spaces are treated as the same. But why is it so....

Like $R^2$ and the set of all ${(x, y, 0) }$ are isomorphic but the "same" vectors in the two spaces are actually different vectors.

Some isomorphic spaces might be having even different rules of vector addition and scalar multiplication, then why the corresponding vectors in both will be the same.

Also any N dimensional vector space $V$ is isomorphic to $F^n$. Buth that n dimensional vector space can be a space of matrices or of polynomials or of any other abstract vectors. How does saying corresponding vectors in each such n dimensional vector spaces are "the same" as the the n tuple in $F^n$.

All these vectors have different rules for multiplication and addition, then what is the intuitive reasoning behind them being treated as same. Will it not defeat the purpose of treating abstract objects as vectors.

Edit:

Precisely this

An n dimensional polynomial space is isomorphic to $F^n$. An n dimensional space of matrices ( n= ab) is isomorphic to $F^n$. Now how is Differentiation in n-dimensional polynomial space mirrored in $F^n$ ( n- tuple are constants) and How is a transpose operation in n dimensional matrix space mirrored to $F^n$. Also since the n dimensional space and n dimensional matrix space are isomorphic to $F^n$ , then they should be isomorphic to each other too ( is this correct). But then how is differentiation in n dimensional polynomial space mirrored to an n dimensional matrix space.

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    $\begingroup$ @ProfessorVector I don't agree with most of what you said. Especially the stuff about people asking for intuition. That is a good thing. Anyways, isomorphic vector spaces, to me, means "the same" vector spaces. This is just a matter of opinion. The opinion here is that the set theoretic definition takes into account the underlying set, while we don't want this. We want to forget this, treat the set elements just as labels and we don't care if someone else labels them differently. $\endgroup$
    – Jens Renders
    Jun 7, 2020 at 20:29
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    $\begingroup$ @ProfessorVector I know that in many cases you do care about the labels, you do care about the underlying set. But in that case I would argue that you simply care about more structure than what a vector space captures. So they are still "the same" as vector spaces, but for example not as topological vector spaces (or sets) or whatever other structure requires you te care about the labels. $\endgroup$
    – Jens Renders
    Jun 7, 2020 at 20:31
  • $\begingroup$ Consider the field $\mathbb{Q}(\sqrt[n]{p})$ for a positive prime $p,$ and the cyclic abelian groups $\mathbb{Z}(n).$ These two objects are the same in a different sense than the one you are concerned with. The lattice of subfields of the one is isomorphic to the lattice of subgroups of the other, So the two algebraic structures, one a field the other not, have the same internal structure as far as the relevant subobjects are concerned. $\endgroup$
    – Chris Leary
    Jun 7, 2020 at 20:32
  • $\begingroup$ A tip to get a good feel for what isomorphisms mean, and why we think of isomorphic structures as the same: Learn basic group theory (and their isomorphisms). Finite groups a more concrete than vector spaces (over R), so it will give you a better feel. $\endgroup$
    – Jens Renders
    Jun 7, 2020 at 20:35
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    $\begingroup$ It might help to think of playing checkers with red and black disks vs. playing checkers with 10 cent and 25 cent coins. Or playing tick-tack-toe with O's and X's vs. playing tick-tack-toe with A's and B's. $\endgroup$
    – Dave L. Renfro
    Jun 9, 2020 at 20:03

6 Answers 6

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You are asking a good question.

Take this statement: For a field $F$, the following vector spaces are isomorphic:

  • $F^{n^2}$
  • The space $M_{n \times n}$ of $n \times n$ matrices over $F$
  • The space $P_{n^2}$ of polynomials with degree less than $n^2$ with coefficients in $F$

The isomorphisms we are talking about in this example only concern the additive structure, the "+", and the scalar multiplication (multiplication with an element in $F$). If we are only allowed to do addition and multiplication with a scalar, then the two spaces behave exactly the same. But you are right that both spaces may allow us to do other things that you cannot naturally do in the respective other space.

But we can always define them in the other space! This is done in general as follows. Take your isomorphism $\phi$, for example $\phi \colon M_{n \times n} \to P_{n^2}$. In $P_{n^2}$ we have differentiation, given by a map $D \colon P_{n^2} \to P_{n^2}$. How can we define differentiation in our matrix space? There is only one way if we want our new definition to be isomorphic to the definition on $P_{n^2}$. We have to define our new differentiation on matrices as $D_M := \phi^{-1} \circ D \circ \phi$. In other words: $$ D_M \colon M_{n \times n} \to M_{n \times n} \\ m \mapsto \phi^{-1}(D(\phi(m))) $$

For example, let's "differentiate" the matrix $$ \pmatrix{ 1 & 2 \\ 3 & 4 } $$

As a polynomial, this is $f(x) = x^3 + 2 x^2 + 3x + 4$ (depends on your choice of $\phi$!). So the derivative is $f'(x) = 0x^3 + 3x^2 + 4x + 3$. As a matrix, this is $$ \pmatrix{ 0 & 3 \\ 4 & 3 } $$

This is your "derivative" of the matrix.

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    $\begingroup$ Thanks for the precise explanation. Now it's very clear to me. I have upvoted it too n will accept it soon enough. Just one thing - 1) instead of writting $ F^ { n^2} \; M _{ n\times n} P _{n^2} $ we could have used $F^{m} , M_{m= n\ times} , P_{m} $ too. This is just notion right. And 2) in your last example your polynomial entries are row wise in the matrix. We could have them column wise too right. You mean exactly that by " depends on your choice of $\phi$, right or is it something else. $\endgroup$
    – Shashaank
    Jun 8, 2020 at 7:08
  • $\begingroup$ The polynomials should have degree at most $n^2-1$, because there's a coefficient for degree $0$ (the constant term). Alternatively you could say "degree less than $n^2$. $\endgroup$
    – celtschk
    Jun 8, 2020 at 7:15
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    $\begingroup$ @Shashaank Thank you. For your second question: Yes, that's correct and that's what I meant by "choice of $\phi$". For your first question: The vector space dimension of the matrix space $M_{m \times n}$ is $m \cdot n$, so the corresponding spaces must be $F^{m \cdot n}$ and $P_{m \cdot n}$. $\endgroup$
    – Vincent
    Jun 8, 2020 at 7:17
  • $\begingroup$ @celtschk Thank you, I will fix it. $\endgroup$
    – Vincent
    Jun 8, 2020 at 7:18
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    $\begingroup$ I really liked this answer. $\endgroup$
    – Zest
    Jun 8, 2020 at 7:19
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Because "isomorphic" literally means "same structure", so thats because two isomorphic spaces are treated as the same. If you think about it, an isomorphism is a bijection with special conditions between the operations in the two different spaces. This basically means that if two spaces are isomorphic, their structure will be the same because the operations work in the same way. In other words, two isomorphic spaces are two different representations of the same structure.

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  • $\begingroup$ How is their structure same. I see only the dimensionality being same. An n dimensional polynomial space is isomorphic to $F^n$ so is an n dimensional matrix space. So the structure of polynomial space should be the same as $F^n$ or the space of n dimensional matrices. But clearly it is not. Isn't it $\endgroup$
    – Shashaank
    Jun 8, 2020 at 4:37
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    $\begingroup$ Well, maybe we should clarify what do we mean when we talk about "structure". Obviously, I was talking about a specific operation that identifies the structure in the single case. For example, let $(A,+_A)$ and $(B,+_B)$ be two groups with the respective $+$ operations. Then they are isomorphic if exists a bijection $f$ such that $f(a_1+_Aa_2)=f(a_1)+_Bf(a_2)$, but we are talking about two groups that are isomorphic regarding + operation. It also could be that $(A,\cdot_A)$ and $(B,\cdot_B)$ are such that $f(a_1 \cdot_A a_2)=f(a_1)\cdot_B f(a_2)$, in this case we are talking about $\cdot$ iso. $\endgroup$
    – Luigi Traino
    Jun 8, 2020 at 7:40
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    $\begingroup$ yes, I got it now. Thank you for the explanation... $\endgroup$
    – Shashaank
    Jun 8, 2020 at 7:42
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Let $V$ be the set of real polynomials of degree at most $1$. Then $V$ is isomorphic to $\mathbb R^2$ under the isomorphism $\phi:ax+b \mapsto (a,b)$.

Does this imply that the elements of $V$ are the same as the elements of $\mathbb R^2$?

Clearly not: $V$ contains functions, $\mathbb R^2$ contains points.

Does this imply that the elements of $V$ behave exactly in the same way as the elements of $\mathbb R^2$?

Yes, their linear properties are the same in the sense that each linear operation in $V$ is mirrored in $\mathbb R^2$ via $\phi$. But not all properties are mirrored: Every non-constant polynomial of degree $1$ has a real zero. This sentence doesn't even make sense in $\mathbb R^2$. But then this sentence is not about linear properties of functions.

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  • $\begingroup$ An n dimensional polynomial space is isomorphic to $F^n$. An n dimensional space of matrices ( n= ab) is isomorphic to $F^n$. Now how is Differentiation in n-dimensional polynomial space mirrored in $F^n$ ( n- tuple are constants) and How is a transpose operation in n dimensional matrix space mirrored to $F^n$. Also since the n dimensional space and n dimensional matrix space are isomorphic to $F^n$ , then they should be isomorphic to each other too ( is this correct). But then how is differentiation in n dimensional polynomial space mirrored to an n dimensional matrix space. $\endgroup$
    – Shashaank
    Jun 8, 2020 at 4:49
  • $\begingroup$ Could you pls let me know this, particularly the last part, regarding differentiation in n dimensional polynomial space being mirrored onto an n dimensional matrix space. $\endgroup$
    – Shashaank
    Jun 8, 2020 at 4:52
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    $\begingroup$ @Shashaank You are asking a good question. The isomorphisms we are talking about in this example only concern the additive structure, the "+". If we are only allowed to do addition, then the two spaces behave exactly the same. But you are right that both spaces may allow us to do other things that you cannot naturally in the respective other space. (But we can always define them in the other space, for example in $\mathbb{R}^2$ we can define the map $(a, b) \mapsto a$ which behaves just like differentiation in the space of polynomials of degree at most $1$). $\endgroup$
    – Vincent
    Jun 8, 2020 at 6:10
  • $\begingroup$ @Vincent .. Thanks, I get your point. This was what I exactly had doubt in. Could you also pls add on to how transposition in the matrix space be mirrored onto $F^n$ ( like you did above) and whether since both the n- dimensional polynomial and n- dimensional matrix space are isomorphic to $F^n$, does it mean they are isomorphic to each other too or not. If they are then like you did can differentiation in n dimensional polynomial space be mirrored onto the n dimensional matrix space too. Maybe of you wish to expand onto your point you made above including this as an answer, I will accept it. $\endgroup$
    – Shashaank
    Jun 8, 2020 at 6:34
  • $\begingroup$ @Shashaank Please see my answer. $\endgroup$
    – Vincent
    Jun 8, 2020 at 6:52
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Consider another case: Integer arithmetic. If humans do it, they typically write the integers in the form of decimal digit strings with an optional sign in front; those digits themselves being patterns drawn on a surface like paper. When computers do it, they represent the integers in binary, and the digits are really different charge states of capacitors somewhere in the computer.

Now decimal digits are something different than binary digits, and patterns drawn on paper definitely are something very different than charge states of capacitors. And yet both the human and the computer will come to the result that multiplying $6$ by $7$ gives $42$. That is, although the differences are vast, they are not relevant for the question of arithmetic (they are of course relevant for other questions, for example if the result will survive a power outage). That is, as far as arithmetic goes, those capacitor states are isomorphic to the patterns drawn on paper.

The same is true for isomorphic vector spaces: As long as you only care about their vector space properties, you don't need to care about whether you have pairs of real numbers, a single complex number, a real function of the form $x\mapsto ax+b$, a translation in the Euclidean plane, or whatever other isomorphic vector space you have. You will always get the very same results.

For example, you will in all cases alike find that you need exactly two basis vectors to span the whole space. And importantly, if you figure out any property in one of the spaces, and it is a property that only refers to the vector space structure, then you will immediately know that it will be exactly the same in all the other isomorphic vector spaces. Just like in the arithmetic example, knowing that in the computer's capacitor-charge representation $6\times 7=42$ means that you also know that if you use the symbols-on-paper representation to work it out, you'll come to the exact same result. Even though in the computer, the $42$ will be represented by the binary digit string $101010$ (or a corresponding pattern of three charged and three uncharged capacitors), and on your paper the same number will be represented by a pattern of lines representing the digit 4 followed by the digit 2.

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Finite dimensional vector spaces are an unusually bad example of a category for learning the lesson that "isomorphic things are the same," because

1) the only isomorphism-invariant of a finite-dimensional vector space is the dimension;

2) in this category there are many examples where two objects are isomorphic, but not canonically, so that really it is not wise to think of them as "the same" without a little extra caution.

Nevertheless, one way to think about this statement is: imagine a sentence that you can write in a formal language, using only $\forall$, $\exists$, 'and', 'or', 'not', and symbols for vector spaces (scalar multiplication, addition, zero, etc.), for example $$ \exists w \in V \ \forall v \in V \ \ \exists c \in \mathbb{R} \ (v + cw = 0). $$

("There is some $w$ in $V$ such that for all $v$ in $V$ there is some scalar $c$ such that $v + cw = 0$.")

Then your sentence will be true in a vector space $V$ if and only if it is true in every vector space that is isomorphic to $V$. (For example, this sentence is true in the zero and one-dimensional vector spaces only). In other words, the truth of all first order sentences is preserved under isomorphism.

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    $\begingroup$ Your mention of canonical isomorphism of course opens another can of worms whether different things sometimes can (and IMHO even should) be treated as "the same for all intents and purposes" $\endgroup$
    – Hagen von Eitzen
    Jun 7, 2020 at 20:18
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    $\begingroup$ The truth of all first order sentences in the appropriate language is preserved ... If $V$ is a two-dimensional vector space, then $\forall f\in V\colon f''(0)=0$ may be true (space of affine linear functions) or false (solutions of $y''+y=0$) or not even make sense (elements of $V$ are not functions to begin with). The problem is of course that "take the second derivative" and "evaluate at $0$" are not in the language of vector spaces $\endgroup$
    – Hagen von Eitzen
    Jun 7, 2020 at 20:24
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Suppose that $\mathbb R^3$ throws a masquerade ball. Everyone puts on a costume, so $(2,3,5)$ looks like $2 + 3x + 5x^2$. Everyone looks different, but secretly everything is the same. It's still the same people and the same relationships. Previously we would say that $(2,3,5) + (1,2,3) = (3,5,8)$. Now, dressed up in costumes, we say that $2 + 3x + 5x^2 + 1 + 2x + 3x^2 = 3 + 5x + 8x^2$. But once you know how to take off the costumes, you see that nothing has changed.

An isomorphism tells you how to take the masks off, revealing that everything is the same.

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  • $\begingroup$ Thanks. This eg cleared many things up. Just one thing. How will differentiation in n dimensional polynomial space be mirrored onto $R^n$ ( considering they are isomorphic) and how will transpose operation in n dimensional matrix space be mirrored onto $F^n$ ( as they are isomorphic). And since both the n dimensional polynomial space and n dimensional matrix space are isomorphic to $F^n$, does it mean they are isomorphic to each other. But then how can differentiation in n dimensional polynomial space be mirrored onto the n dimensional space matrix space. $\endgroup$
    – Shashaank
    Jun 8, 2020 at 6:20
  • $\begingroup$ @Shashaank What do you mean by "$n$-dimensional matrix space"? $\endgroup$
    – littleO
    Jun 8, 2020 at 6:30
  • $\begingroup$ @Shashaank: Differentiation of the polynomial corresponds to the linear transformation denoted by the matrix $$\begin{pmatrix} 0 & 1 \\ & 0 & 2 \\ & & 0 & 3 \\ & & & \ddots & \ddots \\ & & & & 0 & n-1 \\ & & & & & 0\end{pmatrix}$$ $\endgroup$
    – celtschk
    Jun 8, 2020 at 6:37
  • $\begingroup$ @littleO there is theorem that every n dimensional vector space is isomorphic to $F^n$. So the n dimensional matrix space can be seen as $ n = a\times b$ where a and b are the number of rows and columns in the sets of matrices in this space. So the dimensionality is n = ab $\endgroup$
    – Shashaank
    Jun 8, 2020 at 6:39
  • $\begingroup$ @Shashaank: The transpose is just a permutation of vector entries. For example if you identify the matrix $\begin{pmatrix}a & b\\c&d\end{pmatrix}$ with the vector $\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$, then its transpose is the vector $\begin{pmatrix}a\\c\\b\\d\end{pmatrix}$ $\endgroup$
    – celtschk
    Jun 8, 2020 at 6:46

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