# Can all linear operators be represented as matrices? (In infinite dimension.)

Old title: Components of unbounded endomorphism wrt Schauder basis? Changed for discoverarbility.

In finite-dimensional linear algebra, given an endomorphism $$T$$ and a basis $$\boldsymbol{e}_i$$, we can easily find a matrix representation of $$T$$: $$T(v) = T(\sum_i v^i \, \boldsymbol{e}_i) \overset{*}{=} \sum_i T(v^i \, \boldsymbol{e}_i) = \sum_i v^i T(\boldsymbol{e}_i) =: \sum_{i,j} v^i \, T_{ij} \, \boldsymbol{e}_j$$

Let us now assume that $$X$$ is a separable Banach space with a Schauder basis $$\{\boldsymbol{e_i}\}_{i \in \mathbb{N}}$$ and $$T: X \to X$$ is an unbounded linear operator. If we try to replicate the previous equation, the sum will become infinite, and since for unbounded operators $$T(\lim_n v_n) = \lim_n T(v_n)$$ doesn't hold, the equality $$(\overset{*}{=})$$ above doesn't hold.

This would seem to indicate that unbounded operators don't have a coordinate representation in a Schauder basis, at least not in the same sense as bounded operators do. Or is there a simple generalization that I just don't see?

If we now consider $$X$$ to be a separable Hilbert space and $$\boldsymbol{e}_i$$ an orthonormal basis, we know that it's isomorphic to $$\ell_2$$ with basis: $$\begin{pmatrix}1\\0\\0\\\vdots\end{pmatrix}, \begin{pmatrix}0\\1\\0\\\vdots\end{pmatrix}, \begin{pmatrix}0\\0\\1\\\vdots\end{pmatrix}, \;\dots$$ This means that either the endomorphisms on the space of „infinite column vectors“ aren't always „infinite matrices“, or that my previous conclusion about the „non-representability“ of unbounded operators was wrong. Which of these is true?

In summary, these are my questions:

• Is it possible to represent an unbounded operator using its coordinates wrt. a Schauder basis of a separable Banach/Hilbert space?
• If yes, how? If no, are there other practical ways to represent it? (Eg. using Hamel basis? Or as an integral kernel on $$L^p$$?)
• Are all operators on $$\ell_2$$ „infinite matrices“?

The whole field of Analysis is highly dependent on the fact that we may approximate complicated things (think a vector $$x$$ in your space $$X$$) by simpler things (think truncated sums of the representation of $$x$$ in your Schauder basis). These approximations are only useful because the observations we want to make (e.g. $$T(x)$$) respect the approximations we made (i.e., are continuous functions).

Analysis is therefore (almost) completely useless if the quantities we want to measure do not respect approximations.

Therefore, in order to study discontinuous linear transformations you might as well ignore the norm structure of $$X$$ and hence your Schauder basis will be devoid of any interesting properties, beyond being any old linearly independent set. In this case, as you noted, you will need a Hamel basis to describe what you see!

Disclaimer. Many discontinuous linear transformtions may be successfully studied with analytic methods because they sometimes possess hidden continuity properties, such as closed operators.

• Hey Ruy, thank you for your answer. I'm afraid, however, that it doesn't really answer any of my questions. I didn't prove that unbounded operators don't have a representation in Schauder basis – so is it true, or is it not? You also suggest using a Hamel basis, but these are typically not constructible, so they don't use very practical to me – how would I use a Hamel basis to make a „coordinate representation“ of an unbounded operator on, say $L^2$? Thanks.
– m93a
Nov 21, 2020 at 18:06
• If $T:V\to W$ is a linear transformation between vector spaces with Hamel bases $\{e_j\}_j$ and $\{f_i\}_i$, then each $T(e_j)$ may be written as a linear combination $$T(e_j) = \sum_i a_{i,j}f_i,$$ with finitely many nonzero terms. The matrix $\{a_{i,j}\}_{i,j}$ is thus a representation of $T$. Notice that this matrix has finite columns (each column has finitely many nonzero terms) and any such matrix represents an operator. However, if you want to apply this to Banach spaces you will really be in trouble since, as you said, Hamel bases cannot be constructed explicitly!
– Ruy
Nov 21, 2020 at 18:39
• One more point: the unbounded operators on $L^2$ that are relevant to applications are usually only defined on a dense subspace.
– Ruy
Nov 21, 2020 at 18:40
• The best I can do to "prove" that unbounded operators do not have a representation on a given Schauder basis is to note that there are (discontinuous) nonzero operators vanishing on every element of the basis. If you want to represent such an operator using your basis, I believe there is no way other than considering the zero matrix and, of course, this would not say much about your operator!
– Ruy
Nov 21, 2020 at 18:49
• Thank you! I still have a lot of lingering questions, but I guess it will be better to make a new, more specific thread about them.
– m93a
Nov 21, 2020 at 20:19

Claim: Given a Schauder basis $$\{ \boldsymbol{e}_j \} \subset X$$ and a linear operator $$T: \mathrm{D}(T) \subset X \to X$$, the vectors $$T(\boldsymbol{e}_j)$$ aren't always sufficient to describe the action of the operator on a general vector of $$X$$.

Or in other words: There exists a non-zero linear operator, whose matrix is zero.

Proof: Let $$X = L^2([0, 2\pi])$$ be the space of all square-integrable functions equipped with the Fourier basis $$\{ \boldsymbol{e}_j \} = (1,\, \sin x,\, \cos x,\, \sin 2x,\, ...)$$. We now use the distribution theory to construct an unbounded linear operator $$T$$.

Let us take $$\mathrm{D}(T)$$ to be the space of bounded piece-wise continuous functions. Since $$\mathrm{D}(T) \subset L^1_{\mathrm{loc.}}(\mathbb{R})$$, all vectors $$f \in X$$ can be considered regular distributions on $$\mathcal{D}(\mathbb{R})$$ and assigned a distributional derivative $$Df$$. The result will be a sum of a regular function and (singular) delta distributions in discontinuities: $$Df = (Df)_{\mathrm{reg.}} + (Df)_{\mathrm{sing.}}$$. Finally we chose a test function $$\varphi \in \mathcal{D}(\mathbb{R})$$ that is nonzero on $$[0, 2\pi]$$ (ie. $$\varphi(x)=1$$ on $$[0, 2\pi]$$). We define: $$T(f) = (Df)_{\mathrm{sing.}}[\varphi] \; \boldsymbol{e}_1$$ This operator is zero for all continuous functions, including the entire basis $$\{ \boldsymbol{e}_j \}$$, but it is generally nonzero for discontinuous functions.

Corollary: Since, given a Schauder basis, there's a natural isomorphism $$\iota: L^2 \to \ell_2$$: $$\underbrace{f}_{L^2} = \sum_j \underbrace{c_j}_{\ell_2} \boldsymbol{e}_j \: , \quad \iota(f) = \{ c_j \}$$ we can construct a non-zero operator on $$\ell_2$$ whose matrix is only zeros: $$S(a_j) = \iota \, T \, \iota^{-1} \; a_j \quad \neq \quad \sum_{j,k} S_{jk} \, a_j \, \boldsymbol{e}_k = 0$$

Claim: Let $$X$$ be a separable Banach space and $$T: \mathrm{D}(T) \subset X \to X$$ a densely defined closed operator. Given a Schauder basis $$\{ \boldsymbol{e}_j \} \subset \mathrm{D}(T)$$ (such a basis always exists), we construct the coordinates $$T_{jk}$$ of the operator $$T$$: $$T(\boldsymbol{e}_j) =: \sum_k T_{jk} \, \boldsymbol{e}_k$$ Now we define an operator $$T_0: \mathrm{D}(T_0) \to X$$ using those coordinates: $$T_0(v) = T_0( \, \sum_j v^j \boldsymbol{e}_j \, ) = \sum_{j,k} T_{jk} \, v^j \, \boldsymbol{e}_k \: , \qquad \mathrm{D}(T_0) = \{ v \in X \,|\, \text{the sum converges}\}$$ Then $$T_0$$ is also closed and $$T_0 \subset T$$.

Proof: The Schauder basis in question exists, because $$\mathrm{D}(T)$$ is dense in $$X$$ – therefore it is a generating set in the sense of countable linear combinations. We can throw away its linearly dependent vectors until we're left with a basis.

The defining feature of closed operators is: $$T \text{ is closed} \;\wedge\; x_n \to x \;\wedge\; T(x_n) \to y \quad \implies \quad x \in \mathrm{D}(A) \;\wedge\; T(x) = y$$ To show that $$T_0 \subset T$$, we chose an arbitrary $$v \in \mathrm{D}(T)$$ and define: $$x_n = \sum_{j=1}^n v^j \, \boldsymbol{e}_j \: , \quad x_n \to v$$ Then: $$T(x_n) = \sum_{j=1}^n v^j \, T(\boldsymbol{e}_j) = \sum_{j=1}^n \sum_k T_{jk} \, v^j \, \boldsymbol{e}_k$$ If that sum converges for $$n\to\infty$$, by the definition of $$T_0$$, it holds that $$v \in \mathrm{D}(T_0)$$ and the result is equal to $$T_0(v)$$. However, from the closedness criterion, we also know that $$v\in\mathrm{D}(T)$$ and the result is equal to $$T(v)$$. If the sum doesn't converge, by definition $$v\notin\mathrm{D}(T_0)$$, but it doesn't tell us anything about $$T$$. Thus, $$T_0 \subset T$$.

It remains to prove that $$T_0$$ is also closed. We assume $$① \Leftrightarrow x_n \to x$$ and $$② \Leftrightarrow T_0(x_n) \to y$$ and want to prove $$T_0(x) = y$$. If we expand the expressions, we get: $$① \; \implies \; \sum_{j=1}^m (x_n)^j \, \boldsymbol{e}_j \xrightarrow{n} \sum_{j=1}^m x^j \, \boldsymbol{e}_j \\[8pt] ② \; \Longleftrightarrow \; \sum_{j,k} T_{jk} \, (x_n)^j \, \boldsymbol{e}_k \xrightarrow{n} y$$ To simplify the matters, we introduce a double-sequence $$\Phi_{mn}$$: $$\Phi_{mn} = \sum_{j=1}^m \sum_k T_{jk} \, (x_n)^j \, \boldsymbol{e}_k$$ The assumption $$②$$ is equivalent to $$\lim_{n\to\infty} \lim_{m\to\infty} \Phi_{mn} = y$$. The assumption $$①$$ implies that $$\lim_{n \to \infty} \Phi_{mn}$$ exists – as it turns out, the convergence is even uniform wrt. $$m$$ (is it really tho?). Then it follows from the Moore-Osgood Theorem, that $$\lim_{m\to\infty} \lim_{n\to\infty} \Phi_{mn}$$ exists and is equal to $$y$$, too.