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I have a good geometric intuition of why two non-zero vectors $v_1,v_2$ are orthogonal if $<v_1,v_2>$ = 0 (Where $<,>$ denotes the standard dot product in a euclidean vector space).

But I have read (see source bellow) that if we take a symmetrical bilinear form (hence reflexive), then the same concepts of orthogonality apply: If $s(v_1,v_2) = 0$ for non zero vectors, then these are considered to be orthogonal. I don't see how the concept of orthogonality between objects is preserved when taking symmetric bilinear forms. (Which in my understanding are a generalization of dot products, only keeping the linearity in both components in its most general form)

Source: https://en.wikipedia.org/wiki/Bilinear_form#Reflexivity_and_orthogonality

Any clarification would be greatly appreciated!

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  • $\begingroup$ Orthogonality is relative to the form chosen. It is not a fixed thing that all forms preserve. $\endgroup$ – rschwieb May 10 '17 at 17:26
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Generally two vectors are orthogonal with respect to a given inner product $(\cdot,\cdot):V \times V \rightarrow F$ if $(v,w) = 0$, so different vectors, even in the same set, can be orthogonal with respect to different inner products.

The inner product is generally a bilinear form which obeys $(x,y) = \overline{(y,x)}$ (where the line indicates conjugation), but in fields where each element is self-conjugate, the inner product is exactly a symmetric bilinear form. The article argues the converse: that a symmetric bilinear form $B(\cdot,\cdot):V\times V \rightarrow F$ is always an inner product on $V$, thus can be used to establish notions of orthogonality on $V$.

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