I'm having trouble understanding how these two representations can coexist. Citing Wikipedia:

In functional analysis, a branch of mathematics, a finite-rank operator is a bounded linear operator between Banach spaces whose range is finite-dimensional.

Finite-rank operators are matrices (of finite size) transplanted to the infinite-dimensional setting.

$(\dots)$ an operator $T$ of finite rank $n$ takes the form $$Th=\sum _{{i=1}}^{n}\alpha _{i}\langle h,v_{i}\rangle u_{i}\quad {\mbox{for all}}\quad h\in H$$ where $\{u_i\}$ and $\{v_i\}$ are orthonormal bases. $(\dots)$ This can be said to be a canonical form of finite-rank operators.

For $H=\ell^2$ it makes sense but... what if we have, say, $H=L^2([0,1])$ or $L^2(\mathbb{R})$? These are function spaces. How can the range $\operatorname{ran} T$ be finite-dimensional? At the end of the day, it sends functions to functions, and they are infinite-dimensional objects.

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    $\begingroup$ The range of a linear map is a vector space. You can take dimensions of vector spaces. What is the problem here? $\endgroup$ Jan 19 at 15:25
  • $\begingroup$ I'm just confused with the definition of "dimension". I used to look at the number of components of the vector in that space, but here I have functions so... what exactly is a component here? $\endgroup$
    – ric.san
    Jan 19 at 15:35
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    $\begingroup$ That's a terrible way. For example, $\{(\lambda,0): \lambda\in \mathbb{R}\}$ is one-dimensional but vectors here have two components. Am I misunderstanding what you say? $\endgroup$ Jan 19 at 15:37
  • $\begingroup$ No, you're not. So for example $e^{ikx}$ is a basis for $L([0,1])$. What matters is the number of "vectors" $e^{ikx}$. They are actually the "components" with which I construct every other vector (square-integrable function on $[0,1]$). $\endgroup$
    – ric.san
    Jan 19 at 15:44
  • $\begingroup$ What is your definition of basis? Because that is not a Hamel basis (but rather an orthonormal basis). $\endgroup$ Jan 19 at 15:45

1 Answer 1


Dimension is defined for a vector space $V$ as the cardinality of a basis of $V$. In this context it makes no sense to say, a function is an infinite-dimensional object. What is infinite-dimensional is the space of functions $L^2((0,1))$ for example.

In an Hilbert space $H$ we could take a finite-dimensional subspace $U \subset H$ and the projection operator $P$ projecting $H$ onto the subspace $U$. Then $\operatorname{ran} T = U$, so $T$ is a finite rank operator.

If $X,Y$ are Banach spaces we could also take any linear functionals $x_i' \in X'$ and vectors $y_i \in Y$ and define an operator $T: X \to Y$ by $Tx := \sum_{i=1}^n x_i'(x)y_i$. $T$ is a finite rank operator with $\operatorname{ran} T \subset \operatorname{span}(y_i)$, so $\operatorname{dim} \operatorname{ran} T \leq n$.


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