I have been reading some materials and, for the n-th time in my life, there was a definition of an inner product as a function $V \times V \rightarrow F$, where $V$ is an abstract vector space and $F$ is an underlying scalar field.

However, it got me thinking. Inner product is this special function which gives us some number, right? In order to get a number, you must work with numbers.

Now, in general, our abstract vectors are not sequences of numbers, be it matrices, ordered pairs, polynomials or whatever, they are just that, abstract vectors. The only thing that connects these vectors with some numbers are their coordinates with respect to some basis. And in that case, we performed an isomorphism to the vector space $R^n$ (if n is the dimension of the vector space), so we effectively consider only $R^n$ in that case, so let's not do that.

What I am getting here, is how is it possible to construct an inner product on an abstract vector space? How do we take two ordinary abstract vectors and get a number out of it? You can't sum these vectors, you can't multiply them etc., they are not numbers. You can only do that with their coordinates. And if you do that, then you are not defining an inner product on a general vector space, you have defined an inner product on $R^n$ and you are using isomorphism to indirectly define inner product on other vector spaces of the same dimension. That can't be right, can it?

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    $\begingroup$ A vector space admits many possible bilinear forms (the term "bilinear form" is better than "inner product", since the latter term usually involves positivity conditions that make no sense with a general scalar field). There is nothing wrong with using coordinates to construct examples of bilinear forms on a vector space, just as there is nothing wrong with defining groups as specific sets with a suitable law of composition. The main point is that the bilinear form should satisfy some algebraic properties (such as $B(v,w+w') = B(v,w) + B(v,w')$), not whether or not it is defined concretely. $\endgroup$
    – KCd
    Nov 18, 2012 at 8:48
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    $\begingroup$ @Lajka: regarding your last "that can't be right, can it?": why not? $\endgroup$ Nov 18, 2012 at 8:49
  • $\begingroup$ In the vector spaces that I know about, the elements are either "tuples" of some sort, or they are functions. Tuples are not very "abstract", but functions are more so. You can get numbers from functions by integrating them (as in Qiaochu's answer). I'd be interested in seeing an example of a vector space that is even more abstract -- where the members are neither tuples nor functions. Then, if we could define an inner product on this space, we'd scratch the OP's itch (and mine). $\endgroup$
    – bubba
    Nov 18, 2012 at 9:56
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    $\begingroup$ As a rule of thumb: Avoid working with a basis if possible. See Qiachu's example using integration that does not asume a basis. You could come up with sine and cosine functions (which are not even a Hamel basis) and define the same inner product via that basis, but that would be highly unsatisfactory (and simply wrong). $\endgroup$ Nov 18, 2012 at 13:30

2 Answers 2


You can pick a basis and define the inner product by specifying what it does to a basis.

But this is somewhat unsatisfactory. In practice, how you go about writing down a meaningful inner product on $V$ depends on how $V$ itself is constructed. For example, if $V$ is, say, a space of real-valued functions on some measure space $(X, \mu)$, then a natural inner product to write down is

$$\langle f, g \rangle = \int_X f(x) g(x) \, d \mu$$

provided that this integral always converges.


Let $V$ be an arbitrary linear space over $F\in\{\mathbb{R},\mathbb{C}\}$. Consider some Hamel basis of $V$, denote it $\{e_\lambda:\lambda\in\Lambda\}$. Then for each $v\in V$ we have a family of numbers $\{v_\lambda:\lambda\in\Lambda\}\subset F$ such that $$ v=\sum\limits_{\lambda\in\Lambda} v_\lambda e_\lambda\tag{1} $$ Note that only finitely many numbers in $\{v_\lambda:\lambda\in\Lambda\}$ are non-zero, so $(1)$ is well defined. Then you can define inner product by $$ \langle v,w\rangle=\begin{cases}\sum\limits_{\lambda\in\Lambda} v_\lambda w_\lambda &\quad\text{ if }\quad F=\mathbb{R}\\\sum\limits_{\lambda\in\Lambda} v_\lambda \overline{w_\lambda} &\quad\text{ if }\quad F=\mathbb{C} \end{cases}\tag{2} $$ One can check that $\langle\cdot,\cdot \rangle :V\times V\to F$ is well defined inner product.

  • $\begingroup$ Interestingly, if we change the definition to $\langle v,w\rangle=\sum_{\lambda\in\Lambda}v_\lambda w_\lambda e_\lambda^2$ (over reals here), then the inner product possesses a very concrete form --- $\langle v,w\rangle=vw$ --- although we cannot write down the $v_\lambda$s or $w_\lambda$s explicitly. $\endgroup$
    – user1551
    Nov 18, 2012 at 9:27
  • $\begingroup$ @user1551 You can not write $e_\lambda^2$ - this is meaningless, because $e_\lambda$ is an abstract vector, not a number. $\endgroup$
    – Norbert
    Nov 18, 2012 at 9:29
  • $\begingroup$ Oh, typo. I was considering $\mathbb{R}$ as a vector space over $\mathbb{Q}$. $\endgroup$
    – user1551
    Nov 18, 2012 at 9:31

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