This doubt has risen in another thread. I am self-studying linear Algebra by Serge Lang.

Before the proof of this theorem $L_v(w)$ is defined as $L_v=\langle v,w\rangle$ for all $w\in V$."If $L_{v_1+v_2}=L_{v_1}+L_{v_2}$.If $c\in K$, then $L_{cv}=cL_v$"

Theorem: The map $v\to L_v$ of $V$ into $V^{*}$ is an isomorphism.I define $L_v(w)=\langle v,w \rangle$, as the functionals of $V^{*}$

Proof:We know that $\dim V=\dim V^{*}$ and by non-degeneracy of the inner product the kernel of the map is 0.

Theorem: Let V be a finite dimensional vector space over $K$, with a non-degenerate scalar product. Given a functional $L:V\to K$ there exists a unique element $v\in V$ such that: $L(w)=\langle v,w\rangle$

for all $w\in V$.

Proof. Consider the set of all functionals on $V$ which are of type $L_v$, for some $v\in V$. This set is a subspace of $V*$, because of the zero functional is of this type, and we have the formulas


Furthemore, if $\{v_1,...,v_n\}$ is a basis of $V$, then $L_{v_1},...L_{v_n}$ are linearly independent. Proof: If $x_1,...,x_n\in K$ are such that:

$x_1L_{v_1}+...+x_nL_{v_n}=0\\L_{x_1v_1}+...+L_{x_n v_n}=0$

and hence

$L_{x_1v_1+...+x_n v_n}=0$

However, if $v\in V$, and $L_v=0$, then $v=0$ by the definition of non-degeneracy. Hence:

$x_1v_1+...+x_n v_n=0$,

and therefore $x_1=...=x_n=0$, thereby proving our assertion. We conclude that the space of functionals of type $L_v\:(v\in V)$ is a subspace of $V*$, of the same dimension as $V*$, whence equal to $V*$. This proves the theorem.$\blacksquare$ Book: "Linear Algebra" by Serge Lang

In the book it is stated about functionals that "we have the same formalism as with scalar products, except for the fact that in the symbol $\langle\varphi,v\rangle$ the two components do not belong to the same space"

I know examples of functionals like the Dirac functional $\delta:V\to\mathbb{R}$ to be the map such that $\delta(f)=f(0)$, I know also the $ess\sup f$ to be a functional over $\mathscr{M}$ the space of bounded functions. On these last two examples I cannot see a relation between the inner product and respective functionals. Therefore I wonder why the book insists so much on the inner product as being essential for the dual space to be defined. In other thread people insisted that inner products do not imply functionals and there is no need of inner products to have functionals, which seems to me a contradiction of the book.


1) Why does the book uses inner product notation such as $L_v=\langle v,w\rangle$ to define linear functionals?

2) Why does Lang build the basis of dual space this way? Can it be built in alternative way? Considering the fact linear functionals are not inner products.

3) Why is the book right even if inner products are not needed for functionals?

4) Why are not the inner products the same as functionals if both share the same properties?

Thanks in advance!

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Riesz_representation_theorem is the underlying reason. But inner products are not the same as functionals: one is $V \times V \to \mathbb{R}$, the other $V \to \mathbb{R}$. Essentially, this result tells you that an inner product gives you a way to identify $V$ with its dual, which does not exist for general vector spaces. $\endgroup$ – Chappers Jul 9 '17 at 11:04
  • $\begingroup$ @Chappers The name of the chapter I am studying is called "The Dual Space". However the author point out the dual space to be the space of all linear functionals. So are you saying the dual space does not exists for general vector spaces? Does that imply linear functionals do not exist for general vector spaces? $\endgroup$ – Pedro Gomes Jul 9 '17 at 11:09
  • 1
    $\begingroup$ No, the dual space $V^*$ always defined (at least for finite-dimensional spaces), as the space of linear functionals on $V$. The RRT says that if you have an inner-product space (rather than just a vector space), then the dual space is precisely made out of elements of the form $\langle v, \cdot \rangle$ (clearly these are all linear functionals, but RRT says under these circumstances, they are all of them). $\endgroup$ – Chappers Jul 9 '17 at 11:19

Notice this only works for finite dimensional vector spaces, because otherwise the full dual space is not actually isomorphic to the original vector space.

If you want the scalar product to actually be an inner product, this in fact only works for real vector spaces, because the complex inner product is not linear in the second argument, but rather semilinear (this is actually only a convention, and you could make it semilinear in the first argument instead, in which case it would work). It works for a vector space over any field if you define $\langle -,-\rangle$ to be a nondegenerate bilinear form, which is not the same thing.

For a nondegenerate bilinear form over the complex numbers, there are always nonzero vectors $v$ such that $\langle v,v\rangle =0$. What makes it nondegenerate is that there is a $w$ such that $\langle v,w\rangle\neq 0$. This is the important property here. We don't need the form to be symmetric or positive definite, so clearly an inner product is not needed. Over an infinite field there will be infinitely many bilinear forms that do the trick, and they will all give different isomorphisms.

Notice also that just because we define $L_v=\langle v,w\rangle$, this does not mean that $L_v$ is a scalar product, merely that we have used a scalar product to define it. It doesn't make sense to call it a scalar product because it is linear, not bilinear.

Generally if something is used in a proof this does not mean it is necessarily needed, only that it is one possible way to solve the problem. There may be any number of ways that will work that could be completely unrelated.

  • $\begingroup$ But the dual space is isomorphic to $V$, if it is built this way. I cannot think of other way of building the dual space. Could you provide me that? Sorry but I am not understanding you answer. $\endgroup$ – Pedro Gomes Jul 9 '17 at 11:30
  • 1
    $\begingroup$ @Pedro This doesn't define an isomorphism for infinite dimensional vector spaces. The $L_v$ are still linearly independent, but they do not span $V^*$. You would need to take infinite linear combinations. $\endgroup$ – Matt Samuel Jul 9 '17 at 11:32
  • $\begingroup$ I would like your answer to dwell on topics I raised, such as the Dirac functional. I do not understand your point. Do you recognize a vector space over the field $\mathbb{R}$ to have only linear functionals that are inner products? The author also pointed out the hermitian product as being "anti-linear or semi-linear". $\endgroup$ – Pedro Gomes Jul 9 '17 at 12:23

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