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A linear operator on a vector space has a basis through which write its associated matrix. This is certainly true for finite dimensional spaces. But is it still true for infinite dimensional spaces? I can not imagine a matrix written with a base that contains infinite elements. How should it be this matrix? With endless columns and endless rows?

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Yes. A linear map between two $K$-vector spaces $V$ and $W$ with given bases can be represented as a matrix with respect to these bases. When the spaces are infinite-dimensional, this will indeed be a matrix with "endless" columns and "endless" rows. (But every column has only finitely many nonzero entries, because linear combinations are finite.) A matrix (in a sufficiently general meaning of this word) over the field $K$ is just a family $\left(a_{i,j}\right)_{\left(i,j\right)\in I\times J}$ of elements of $K$ indexed by a product $I \times J$ of two sets $I$ and $J$. The matrices usually seen in linear algebra books have $I = \left\{1,2,\ldots,n\right\}$ and $J = \left\{1,2,\ldots,m\right\}$, but there is nothing wrong with allowing arbitrary $I$ and $J$.

For instance, consider the ring $K\left[x\right]$ consisting of all polynomials in the variable $x$ over our ground field $K$. The matrix which represents the map $K\left[x\right] \to K\left[x\right], \ p \mapsto p'$ (where $p'$ means the derivative of $p$) with respect to the basis $\left(1,x,x^2,\ldots\right)$ of the $K$-vector space $K\left[x\right]$ has the form

$\left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 2 & 0 & \cdots \\ 0 & 0 & 0 & 3 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots\end{array}\right)$.

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  • $\begingroup$ Unless my proof is incorrect (which is always possible), there are operators which cannot be (faithfully) represented by an infinite matrix. See: math.stackexchange.com/a/3918826/142487 $\endgroup$
    – m93a
    Aug 29, 2021 at 20:28
  • $\begingroup$ @m93a: A Schauder basis is not a basis in the original meaning of the word. $\endgroup$ Aug 29, 2021 at 20:36
  • $\begingroup$ If one works with Hilbert spaces, the word basis is always understood as a Schauder basis, not a Hamel basis. I think it depends on the context (which isn't really clear from the question). However, you are of course right that with Hamel basis, linear operators always can be represented by “matrices” (with uncountably many elements though). $\endgroup$
    – m93a
    Aug 29, 2021 at 20:56

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