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I was curious about how to define a linear function in a vector space whose basis is not countable. In this case, to have a concrete example, I though about the sequence space in $\mathbb{R}$, that is,

$$\mathbb{R}^\mathbb{N}=\prod^{\infty}\mathbb{R}$$

Imagine that we wanted to define a linear function $A_{i,j}$ that for a sequence $x$ in $\mathbb{R}$ it acts as a swap elements operator:

$$x=(x_1, x_2, ..., x_i, ..., x_j,...)$$

$$A_{i,j}(x) = (x_1, x_2, ..., x_j, ..., x_i,...)$$

If the basis is not countable, how can we express $x$ in terms of a linear combination of the elements of the basis $\{e_1, e_2, ...\}$ i.e.: as a vector of some coordinates? And, without this notation, how can we express our function $A_{i,j}$? In the case of $\mathbb{R}^n$, it is be trivial to construct a swap matrix that represents such function. Is it even possible to find a (infinite) matrix form in this other case?

Edit: thank you very much for the comments. One of them made me realise that maybe my choice of the function $A_{i,j}$ is not the best for the question I have, since it only acts on a finite dimensional subspsce of $\mathbb{R}^\mathbb{N}$. In this sense, a better example could be a function $B_i$ that adds a $0$ at the i-th position of a sequence, that is,

$$B_i(x) = (x_1, x_2, ..., x_{i-1}, 0, x_i,...)$$

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    $\begingroup$ Linearity is just defined as $f(\lambda x+\mu y)=\lambda f(x)+\mu f(y)$. For the finite dimensional case, you probably prove in linear algebra that they coincide with matrices. $\endgroup$
    – Vincent Batens
    Commented Feb 21 at 22:22
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    $\begingroup$ On another note, why is expressing $x$ as a linear combination a problem? After all, by definition every element of a vector space is a finite linear combination of basis elements. I suspect a lot of confusion here arises from the fact that you have not found a basis (although this might prove to be difficult). $\endgroup$
    – Ben Steffan
    Commented Feb 21 at 22:24
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    $\begingroup$ @BenSteffan That's a great point I was missing, thank you. Nevertheless, I unfortunately still don't see how that can help in finding a way to represent an arbitrary linear function such as the proposed $A_i,j$ in this space $\endgroup$
    – user3141592
    Commented Feb 21 at 22:27
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    $\begingroup$ It's not totally clear to me what you want to achieve. What do you mean by "in a way that we can work with them" and "how can we express our function $A_{i,j}$"? You have already expressed it, in a way that's much clearer than writing down an infinite matrix! Can you give an example of a problem that you think could be solved by finding a better representation? Typical applications of matrices in finite dimensions are things like calculating determinants or inverses, but in infinite dimensions those are pretty hopeless in general. $\endgroup$
    – Izaak van Dongen
    Commented Feb 21 at 22:35
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    $\begingroup$ I note that any such $A_{i,j}$ only acts on the $\mathbb{R}^{\max(i,j)}\ $ subspace of $\mathbb{R}^{\mathbb{N}}$. So, perhaps it is smarter to think of $A_{i,j}$ as the identity except on that subspace and then write the matrix on that subspace (using a $\max(i,j)$-element basis for that subspace)? Of course, one can then make this "stupid" by (1) pick $A_{i,j}$, (2) pick an infinite basis starting with $\hat{x}_i, \hat{x}_j, \dots$ and then (3) have the resulting matrix be the two dimensional swap. $\endgroup$
    – Eric Towers
    Commented Feb 22 at 1:09

2 Answers 2

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It is not (usually) required that it is possible to represent a linear function from one vector space $V$ to $W$ as a matrix. It just needs to have these two properties:

$$f(u+v) = f(u) + f(v)$$

$$f(\lambda v) = \lambda f(v)$$

These can be compressed into one rule as in Vincent Batens' comment.

For a finite dimension space and some countable infinite cases, it is possible to represent the linear function as a matrix but this is not required. An arbitrary uncountable dimension vector space only has a basis if we accept the Axiom of Choice.

An interesting example of a linear function which cannot easily be represented as a matrix is the Fourier transform.

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Note that your $B_i$ is not linear, because $B_i(x+y)\not = B_i(x) + B_i(y)$.

Differentiation over the space of all differential functions is one common example of a linear map over an infinite dimensional vector space.

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  • $\begingroup$ I edited a typo in the definition of $B_i$ to solve the issue with linearity. This should have been posted as a comment instead of as an answer, since it is not directly addressing the scope of the question $\endgroup$
    – user3141592
    Commented Feb 22 at 15:24
  • $\begingroup$ Are you sure you meant to ask for an uncountable space? $B_i$ only makes sense if the sequence is countable. $\endgroup$
    – Xodarap
    Commented Feb 22 at 21:27
  • $\begingroup$ why is that? note that $B_i$ is not performing a change of coordinates of a vector in our vector space, but of elements in our infinite sequences $\endgroup$
    – user3141592
    Commented Feb 22 at 21:54

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