I just happened across this over on Math Overflow. It references the following theorem from linear algebra:

A vector space has the same dimension as its dual if and only if it is finite dimensional.

I would like to ask a physical question using the infinite square well (ISW) in quantum mechanics as motivation. For the ISW we obtain $$\psi_n=A_n\sin(\frac{n\pi x}{a})$$ as the eigenfunctions of the Hamiltonian. Here $n=1,2,3,4...$ enumerates the states. If I understand correctly this is an infinite dimensional vector space, because the $\psi_n$'s form an infinitely large basis (ie there is no largest value of $n$). If the dual space is the set of functions $\psi_n^*$ (which I think it is) how can the vector space and the dual space have different dimensions?

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    $\begingroup$ Can you cite the post? There is more to this than what you say: the "theorem" as it stands is wrong: any separable and infinite dimensional Hilbert space being a counterexample, if "same dimension" is to be read as "having cardinally equivalent basis sets" and if dual is to be read as "topological dual" (i.e. set of all cts linear functionals on the space) $\endgroup$ Commented Dec 14, 2014 at 8:20
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    $\begingroup$ They are referring to the "algebraic dual", i.e. the set of all linear maps (not necessarily continuous) from the vector space to the field where it is defined. This is a larger set than the topological dual. The Riesz's Representation Theorem holds only for topological dual pairs. $\endgroup$
    – yuggib
    Commented Dec 14, 2014 at 9:34
  • $\begingroup$ @yuggib That's what I wanted to make clear. Anyhow, my answer here may be relevant $\endgroup$ Commented Dec 14, 2014 at 12:29
  • $\begingroup$ Also note that the $\psi_n$ of your example do not form a basis in the algebraic sense (which is the context of the theorem): a basis is a linearly algebraic subset such that every element can be written as a finite linear combination. The cardinality of a basis in this sense is the same for all bases and is by definition the dimension. $\endgroup$
    – doetoe
    Commented Dec 14, 2014 at 23:37

2 Answers 2


There are two concepts of duality for vector spaces.

One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The proof you can see on math overflow uses, roughly speaking, the fact that the cardinality of $\mathbb{K}^V$ is strictly larger than the cardinality of $\mathbb{K}$ if $V$ is infinite dimensional and has at least the same cardinality as $\mathbb{K}$.

So for algebraic duals, the dual of any infinite vector space has bigger dimension than the original space.

The other concept is the topological dual, that can be defined only on topological vector spaces (because a notion of continuity is needed). Given a topological vector space $T$, the topological dual $T_{top}^*$ is the set of all continuous linear functionals (continuous w.r.t. the topology of $T$). It is a proper subset of the algebraic dual, i.e. $T_{top}^*\subset T_{alg}^*$.

For topological duals, the restriction to continuous functionals makes the previous statement false (i.e. there exist infinite dimensional topological vector spaces whose topological dual has the same dimension of the original space).

The usual example are Hilbert spaces, where the Riesz representation theorem holds (see my comment above): any object of the topological dual $H^*_{top}$ of a Hilbert space $H$ can be identified via isomorphism with an element of $H$. So an Hilbert space and its dual are the "same".

Note however that the topological dual is always thought to be "bigger (or maybe equal)" than the original space. I am very non-precise here, but I think the following example clarifies. Think to the distributions $\mathscr{S}'(\mathbb{R})$. This is the topological dual of the functions of rapid decrease $\mathscr{S}(\mathbb{R})$. Any $f\in \mathscr{S}$ is isomorphic to a distribution in $\mathscr{S}'$, but the converse is obviously not true: there are distributions that are not functions (the Dirac's delta), and in general any $L^p$-space is thought as a subset of $\mathscr{S}'$ (so $\mathscr{S}'$ is quite "big").

  • $\begingroup$ The topological dual is not always bigger the original space. For example, the strong topological dual of $\mathscr{S}'(\mathbb{R})$ is again $\mathscr{S}(\mathbb{R})$. These spaces are reflexive. $\endgroup$ Commented Nov 12, 2021 at 19:14

Restricting ourselves to just vector spaces without any extra structure, the theorem is true.

One way to see this is to note that any member $f$ of the dual space is uniquely defined by the value it returns acting on the basis $\{\psi_n\}$, say $f(\psi_n) = z_n$ for complex numbers $z_n$. Then $V^*$ is isomorphic to $\mathbb{C}^\mathbb{N}$, the set of sequences of complex numbers. It is a well-known fact that $\mathbb{R}^\mathbb{N}$ does not have a countable basis as a vector space over $\mathbb{R}$, and it is a simple matter to extend this to $\mathbb{C}^\mathbb{N}$ not having a countable basis over $\mathbb{C}$. If this doesn't seem intuitive (e.g. you jump to thinking of the "basis" $\{(1,0,0,\ldots), (0,1,0,\ldots), \ldots\}$), the key is that only finite sums are allowed in raw vector spaces; what would it even mean to add an infinite number of vectors without a notion of convergence?

A more physically-inspired argument against the idea that complex conjugation yields (a basis for) all members of $V^*$ is to consider delta-functions. For some $x_0$ in the interval, consider "$\delta(x-x_0)$," the "function" that integrates against $f \in V$ to return $f(x_0)$. In actuality, $\delta$ is a perfectly valid member of $V^*$, defined by $\delta(\psi_n) = \psi_n(x_0)$. Suppose $\delta = a_1 \psi_{n_1}^* + \cdots + a_k \psi_{n_k}^*$ could be written. But then $a_1^* \psi_{n_1} + \cdots + a_k^* \psi_{n_k}$ would be a perfectly well-behaved finite sum of sines that was the complex conjugate of the delta "function" -- a clearly nonsensical result. Besides, $$ (a_1 \psi_{n_1}^* + \cdots + a_k \psi_{n_k}^*)(\psi_n) = a_k \delta_{n,n_k}, $$ which is $0$ for all but finitely many $n$, whereas $\psi_n(x_0)$ can be nonzero for every $n$ (choose $x_0$ to be an irrational multiple of $a$).


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