I read that given a vector space $V$, and two vectors in $V$, then the two vectors may be orthogonal under one inner product definition but not orthogonal under a different inner product. Considering that, would it be fair to say that one of the purposes of defining an inner product is to define what it means for two vectors to be perpendicular?

Separately, are there any definitions of inner product other than the standard Euclidean inner product in $\mathbb{R}^n$ that are widely used in practical applications? I guess that would that be equivalent to asking if there are alternative definitions of orthoginality which are practically useful?

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    $\begingroup$ I always think of the inner product intuitively as measuring "angles between vectors", although this is not a valid interpretation in every case. So not only does it determine orthogonality, it determines the range of angles. $\endgroup$
    – rschwieb
    Jun 27, 2012 at 10:22
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    $\begingroup$ There is a subtle difference between fixing an inner product and fixing a notion of orthogonality: the latter does not "see" a positive scalar factor by which the inner product might be multiplied. $\endgroup$ Jun 27, 2012 at 13:14
  • $\begingroup$ @MarcvanLeeuwen You mean like a weighted Euclidean inner product? $\endgroup$ Jun 27, 2012 at 17:11

3 Answers 3


Knowing the set of orthogonal pairs of vectors fixes an inner product up to a constant positive factor. It is clear that such a factor doesn't change the concept of orthogonality. Conversely, we can prove that two inner products with the same set of orthogonal pairs must be related through a constant factor:

Given two inner products $\langle -,-\rangle$ and $[-,-]$ such that they agree about which vectors are orthogonal. Let $v_1,\ldots,v_n$ be an orthonormal basis with respect to $\langle -,-\rangle$. By assumption $[-,-]$ agrees that the $v_i$s are mutually orthogonal, so knowing the value of $[v_i,v_i]$ for each $i$ will fix all of $[-,-]$ by linearity.

Now for $i\ne j$ we have $$\langle v_i+v_j, v_i-v_j\rangle = \langle v_i,v_i\rangle - \langle v_j,v_j\rangle = 1 - 1 = 0 $$ and therefore it must also hold that $[v_i,v_i]-[v_j,v_j] = [v_i+v_j, v_i-v_j] = 0$. Since $i$ and $j$ were arbitrary, all the $[v_i,v_i]$s must be the same, so $[v,w]=a\langle v,w\rangle$ for all $v$, $w$, where $a=[v_1,v_1]$.

If the vector space is infinite-dimensional (such that there is not necessarily any orthogonal Hamel basis for all of it), this argument can be repeated for each finite-dimensional subspace to reach the same conclusion.

As for the second question, the Gram--Schmidt process shows that every inner product on $\mathbb R^n$ is the same, modulo changes of basis. But in situations where we have a preferred basis imposed on us externally it certainly makes sense to consider different inner products. The most intuitively vivid examples come from differential geometry where "custom" inner products are used to connect arbitrary not-necessarily-rectilinear coordinate systems with geometric reality. For example, consider geographic coordinates on the Earth. If we have two lines on the map (of a not too large area) given by coordinates in degrees, then it makes sense to ask for the angle between them. We can get that using an inner product -- but because a degree of longitude is shorter than a degree of latitude (except at the equator) this needs to be a non-standard inner product in order to give geometrically meaningful results.


Specifying an alternative inner product on $\mathbb R^n$ amounts to specifying a positive definite symmetric $n\times n$ matrix (whose entries specify the new inner products between the standard basis vectors). Conversely most uses of symmetric matrices can be interpreted as specifying a symmetric bilinear form (which is an inner product if the matrix is positive definite) that has some relevance to the problem at hand, or more precisely as expressing such an bilinear form in terms of another (standard) one. Given the pervasiveness of symmetric matrices in (notably) applied mathematics, I don't doubt that the business of comparing different inner products has great practical importance.


An inner product basically generalises the notion of "dot product". It allows you to define generalised angles between vectors and generalised lengths (without having to define these concepts first).

See my answer here:


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    $\begingroup$ This should have been a comment, or more elaborated here. If you want to call this an answer at least give a summary of your argument in the link. $\endgroup$
    – Asaf Karagila
    Jun 27, 2012 at 10:41
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    $\begingroup$ But this is a valid answer, the OP is going to follow the link anyway and read my other answer...so why repeat any of it? $\endgroup$
    – fretty
    Jun 27, 2012 at 10:59
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    $\begingroup$ The wish is granted though...there are some strange customs on this site! $\endgroup$
    – fretty
    Jun 27, 2012 at 11:04
  • $\begingroup$ I took a look at the link, but I have no idea what the Grahm-Schmidt procedure is. $\endgroup$ Jun 27, 2012 at 11:06
  • $\begingroup$ What do you mean by generalized angles and lengths? $\endgroup$ Jun 27, 2012 at 11:09

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