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.
Best Answer
You are asking a good question.
Take this statement: For a field $F$, the following vector spaces are isomorphic:
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.