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Any linear map between two finite-dimensional vector spaces can be represented as a matrix under the bases of the two spaces.

But if one or all of the vector spaces is infinite dimensional, is the linear map still represented as a matrix under their bases?

If there is matrix of infinite dimension, what is it used for if not used as a representation of a linear map between vector spaces?

Thanks and regards!

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2 Answers 2

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Sure, if $T:V\to W$ is a linear transformation between vector spaces $V$ and $W$ with bases $B$ and $C$, respectively, then $T$ can be described in terms of the coordinates with respect to these bases, thus yielding a "matrix". How closely this relates to the usual notion of matrix depends on the nature of $B$ and $C$. In the usual notion, you take bases that are not only finite, but ordered, so that it makes sense to talk about the 1st row, etc., of the matrix; that is, you make all bases indexed by sets of the form $\{1,2,\ldots,n\}$. The closest to this in the infinite dimensional setting would be to have bases indexed by the positive integers.

More generally, suppose $B=\{v_j\}_{j\in J}$ and $C=\{w_i\}_{i\in I}$, where $I$ and $J$ are sets. Then the matrix of $T$ can be described as a function $M:I\times J\to F$, where $F$ is the base field, by taking $M(i,j)$ to be the coefficient of $w_i$ in the $C$-expansion of $Tv_j$. Such matrices are column-finite, in the sense that for each $j\in J$, the set of $i\in I$ such that $M(i,j)\neq0$ is finite. Conversely, each column-finite matrix, in this sense, corresponds uniquely to a linear transformation between $V$ and $W$. Coordinate-wise addition of such matrices corresponds to addition of the linear transformations.

You can also extend multiplication. Suppose that $S:W\to X$ is a linear transformation and that $X$ has basis $D=\{x\_k\}\_{k\in K}$. Let $N:K\times I\to F$ denote the $C$-$D$ matrix of $S$. Then $ST:V\to X$ has $B$-$D$ matrix $NM:K\times J\to F$ defined by $$(NM)(k,j)=\sum_{i\in I}N(k,i)M(i,j).$$ In particular, note that this sum is always finite because $M$ is column-finite.


Motivated by Calle's answer, I decided to add a little on a different kind of matrix for continuous linear transformations on Banach spaces with Schauder bases.

If $X$ is an infinite dimensional separable Banach space, then a sequence $(e_n)_{n=1}^\infty$ in $X$ is called a Schauder basis for $X$ if every $x\in X$ has a unique representation $x=\sum_{n=1}^\infty a_ne_n$, the $a_n$ being scalar and the sum being norm convergent. If $X$ and $Y$ are Banach spaces with Schauder bases $(e_n)$ and $(f_n)$ respectively, and if $T:X\to Y$ is a bounded linear operator, then $T$ can be described by a matrix $(a_{ij})_{i,j=1}^\infty$, with $a_{ij}$ being the coefficient of $f_i$ in the $(f_n)$ expansion of $Te_j$. The map from bounded operators to matrices in one-to-one and preserves algebraic structure, but there is typically not any nice description of which matrices correspond to bounded operators.

For example, in a separable Hilbert space any orthonormal basis is a Schauder basis. For maps between Hilbert spaces the coefficients are found as $a_{ij}=\langle Te_j,f_i\rangle$. In $c_0$, the space of sequences converging to $0$ with sup norm, and in $\ell^p$, the sequence space with norm $\|(x_n)_{n=1}^\infty\|_p=(\sum_{n=1}^\infty|x_n|^p)^{1/p}$, the sequence $(e_n)$ such that the $n^\text{th}$ component of $e_n$ is $1$ and all other components are $0$ forms a Schauder basis.

If $c$ is the space of convergent sequences with sup norm, then this will no longer be a Schauder basis, and in particular it is clear that $\sum_{n=1}^\infty x_n e_n$ is not norm convergent unless $\lim_{n\to\infty}x_n=0$. A Schauder basis for $c$ can be obtained by adding $e_0=(1,1,1,\ldots)$. If $(x_n)\in c$ and $x=\lim_n x_n$, then $(x_n)=xe_0 +\sum_{n=1}^\infty(x_n-x)e_n$ is the basis representation. As in Calle's answer, suppose that $T:c\to c$ is defined by $T(x_1,x_2,x_3,\ldots)=(x,0,0,\ldots)$. Then $T$ has a matrix representation with repsect to $(e_0,e_1,\ldots)$ (but not with respect to $(e_1,e_2,\ldots)$), namely $a_{10}=1$ and all other components are $0$.

Similar to Olod's warning, such matrices typically play only a marginal role, even in cases where they are guaranteed to exist, like on Hilbert space. Not every separable Banach space has a Schauder basis. Enflo first gave an example of a separable Banach space without the approximation property, which guarantees that it has no Schauder basis.

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Thank you for pointing this out. Can you give an example where $\sum_{n=1}^\infty x_n e_n$ is not norm convergent in $c$? For example, if I write $e_0$ in this basis, it seems to me that the sequence $\|\sum_{n=1}^m x_n e_n\| = \sup_{i} x_i = 1$, since all $x_i = 1$. What am I missing? –  Calle Jan 20 '11 at 4:16
    
@Calle: $\|e_0-\sum_{n=1}^m e_n\|=1\not\to0$. –  Jonas Meyer Jan 20 '11 at 4:24
    
Ah, yes, of course. Thank you. But it would still be a Hamel basis, right? –  Calle Jan 20 '11 at 4:31
    
@Calle: It is far from a Hamel basis. Every vector must be a finite linear combination of elements of a Hamel basis. For example, $(1,1/2,1/3,1/4,\ldots)$ is not in the span of $\{e_n\}_n$. A Hamel basis of an infinite dimensional Banach space has cardinality at least $2^{\aleph_0}$ (jstor.org/stable/2318458), and it is consistent with ZF that no infinite dimensional Banach space has a Hamel basis (mathoverflow.net/questions/5303/basis-of-linfinity/5313#5313). –  Jonas Meyer Jan 20 '11 at 4:51
    
Okay. Thank you for pointing out my faulty reasoning. Do you mind putting my argument in a quote or something in your post so I can delete my post? –  Calle Jan 20 '11 at 5:33
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What, however, must be understood that the role of matrices when one works with linear operators of infinite-dimensional vector spaces is a (very) marginal one. Familiar techniques such as the use of determinants, traces etc. no longer work. For instance, any determinant-like function on $\mathrm{GL}(V)$ where $V$ is an infinite-dimensional vector space over a field is necessarily the trivial one, since any element of $\mathrm{GL}(V)$ is a product of commutators (the group $\mathrm{GL}(V)$ is perfect, in other words; a result by Alex Rosenberg of 1958.)

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