Let's look at the well-known definition of orthogonal vectors:

Let $V$ be a vector space. Two vectors $x, y \in V$ are orthogonal to each other when the following condition is fulfilled: $$\langle x,y \rangle = 0$$

Let me explain, where I see a contradiction in this definition.

As I understand the inner product $\langle\cdot \rangle$ of two vectors is not unambiguous (like a norm of a vector, for instance). It can be defined in many ways for different vector spaces, it just needs to satisfy some properties.

As inner product can be defined differently, it can presumably take different values for the same two vectors, depending on which 'version' of the inner product is applied to these vectors. Just analogically with the definition of the norm. Let me illustrate this on an example:

Consider a vector $x=(1,-3,2)^T$. Its Euclidean norm is $\lVert x\rVert_2=\sqrt {14} = 3.74\dots$, its Maximum norm is $\lVert x\rVert_\infty = \max(\lvert 1 \rvert, \lvert -3 \rvert, \lvert 2 \rvert) =3.$ Clearly these two norms are not equal.

So I guess the same generalization is applied to the definition of the inner product: for some vector space it can be defined in many different ways and thus can take different values upon the applied definition. So theoretically in some special case it might happen so, that the value of an inner product of two vectors, that was defined in one way can be equal $0$ which will imply that the vectors are orthogonal, while the value of an inner product defined in another way applied to the same two vectors will not be equal $0$, which will imply that the vectors are not orthogonal. There's an obvious contradiction here.

There's no such contradiction in the definition of norm as there's a theorem that specifies equivalence of two norms:

For some finite-dimensional vector space $V$ let $\lVert x\rVert$ and $\lVert x \rVert '$, $x\in V$ be two different norms. Then $$\exists c \ge 0 \; \forall x \in V: \quad \frac 1c \lVert x\rVert ' \leq \lVert x\rVert \leq c\lVert x\rVert '$$

Clearly when some norm evaluates for some definite vector to $0$ (thus the vector has 'zero length') any other norm will evaluate to $0$ according to the equivalence theorem. So it cannot happen so, that a zero vector can have a 'non-zero length'.

Therefore the one way to solve this contradiction is to prove that for some vector space $V$ an inner product is defined unambiguously. Another way would be to set an analogical theorem for inner products stating their equivalence, from which will follow, that if some inner product is evaluated to zero, all other will do so. Let's again consider Euclidean space, for which the inner product (also called dot product) is defined as $$\langle x,y \rangle = a_xa_y + b_xb_y,$$ where $a_i$ and $b_i$ are the corresponding coordinates of the vectors $x$ and $y$. This is really an inner product as it fulfills the properties. However has it been proved that in the Euclidean space there can't exist any other prescription, which, when applied to two vectors will fulfill the properties of the inner product and thus be another type of the inner product for this space? Or does some theorem exist that states equivalence of two inner products?

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    $\begingroup$ Are you objecting that different choices of inner product for some set of vectors can lead to two vectors being orthogonal with respect to one inner product but not the other? I don't see any immediate problem with that. $\endgroup$ – Semiclassical Jul 30 '14 at 15:27
  • $\begingroup$ @Semiclassical yes, you got it right $\endgroup$ – Dmitry Kazakov Jul 30 '14 at 15:30
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    $\begingroup$ @Semiclassical so orthogonality is not an absolute propety, it is made with respect to an inner product, isn't it? $\endgroup$ – Dmitry Kazakov Jul 30 '14 at 15:34
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    $\begingroup$ Right. Though, as per my comments in martini's answer, I wonder if there's something analogous to the equivalence theorem you cite (though of course it would have to be a weaker result.) $\endgroup$ – Semiclassical Jul 30 '14 at 15:37

No. Your definition of orthogonaly is not fully correct, as you write correctly. Orthogonality only makes sense with respect to a given inner product, not for an arbitrary vector space. It should be written as:

Let $(V,\left<\cdot\right>)$ be an inner product space. Two vectors $x,y\in V$ are orthogonal to each other when the following condition is fulfilled: $$\left<x,y\right> =0 $$

An inner product space is an vector space with a fixed inner product. Orthogonality depends on the inner product. On the spaces $\mathbb R^n$ and $\mathbb C^n$ one often drops the mentioning of the inner product, in this case one always wants the "standard" inner product on these spaces, namely the euclidian (unitarian) one.

To give an example, consider $\mathbb R^2$ with the inner product $$ \left<x,y\right> = (x_1 - x_2)(y_1 - y_2) + x_2y_2 $$ This is an inner product, and $(1,1)$ and $(1,0)$ are orthogonal with respect to it (but not for the euclidian inner product).

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  • $\begingroup$ I'd be interested in how different (loosely speaking) two inner products on the same space can be with respect to orthogonality. (Based on that example, I'd suspect the answer is 'as different as you like.') $\endgroup$ – Semiclassical Jul 30 '14 at 15:35
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    $\begingroup$ Yes. For any invertible matrix $A$ on $\mathbb R^n$, $\left<x,y\right> := \sum_{i=1}^n (Ax)_i(Ay)_i$ is an inner product. $\endgroup$ – martini Jul 30 '14 at 15:36
  • $\begingroup$ This is a notable extension to my version of definition, now it is clear $\endgroup$ – Dmitry Kazakov Jul 30 '14 at 15:37

different inner products establish different geometries , so orthogonality may changes. Every real inner product on a real vector with dimension $m$ can be determined by a positive symmetric $m\times m\ $ real matrix $M$ in this way $$ <X,Y>=X^{T}MY $$ and conversely every $$ X^{T}MY $$ which $ M$ is a positive symmetric $m\times m\ $ real matrix is an real inner product.

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