# Intuition for orthogonal vectors in $\Bbb R^n$

Two vectors in $\Bbb R^n$ are orthogonal iff their dot product is $0$. I'm aware that the dot product can be defined in other spaces, but to keep things simple let's restrict ourselves to $\Bbb R^n$.

Given that the idea of orthogonality is roughly to identify when two vectors have no "overlap", then apart from the fact that in $\Bbb R^2$ and $\Bbb R^3$ this corresponds to the geometrical notion of orthogonality, why is this chosen as the definition of orthogonality? Ideally give examples of concrete mathematical problems where this definition arises naturally.

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It does not only correspond to the geometrical definition of orthoganilty in 2 and 3 dimensions, but in all dimensions. – Raskolnikov Jul 14 '13 at 8:20
@Raskolnikov What do you mean? – Jack M Jul 14 '13 at 10:07
How is the geometrical definition of orthogonality different in 10 dimensions from the one in 3 dimensions? – Raskolnikov Jul 14 '13 at 11:03
This question is virtually impossible to answer, because it is not clear what is being asked, and you present a confused idea of what orthogonality is about and seem to ask for help progressing in this confused view. Notably "having no overlap" for vector has neither any formal nor any intuitive meaning. You could be asking one of (1) why should orthogonality involve a notion of inner product (2) what properties should an inner product have intuitively (3) why does the dot product satisfy our intuitive requirements of an inner product, and in particular what we expect of orthogonality? – Marc van Leeuwen Jul 14 '13 at 13:46
@MarcvanLeeuwen But "having no overlap" is how Wikipedia introduces the concept of orthogonality almost verbatim. – Jack M Jul 14 '13 at 18:42

Consider two independent vectors in $\mathbb R^n$. Then there's a unique plane through them (and the origin). Pick a linear transformation from $\mathbb R^n\to\mathbb R^2$ that sends that plane onto all of $\mathbb R^2$ and preserves lengths. Then the vectors are dot-product-orthogonal in $\mathbb R^n$ if and only if their images are geometrically-orthogonal in $\mathbb R^2$ (use the converse of Pythagorean theorem or whichever other criterion you like). This says that the geometry in $\mathbb R^n$ really acts the same way as in $\mathbb R^2$.

However, you seemed to be saying that you were not interested in the geometry side of things, and wanted an independent reason why orthogonality of vectors is important. Pretty much everything in your linear algebra book about orthogonality is an example, with one of the biggest ones being the spectral theorem, which says that every nice linear map (given symmetric matrices in the $\mathbb R^n$ case) is just a linear combination of orthogonal projections.

Another important use of orthogonality is in the singular value decomposition, which has many applications, including image compression and topography.

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Orthoganility intuitively could be reflcted as "no dependency". It is deeply anchored in the intuition of eigen-space and eigen-vector and eigen-value, eigen-... Recall that eigen means in German "the very geniune property of itself" and this relates to a property which is absolutely genuine to $A$ and not shared with any $B$. So keeping also in mind that orthogonality of eigen-properties is relative. Indeed in $\Bbb R^n$ it ties depply with the dimmensions. There, dimmensions represent the number of possible eigen-properties (metaphoric: absolute selfish properties). The term orthogonality is lent indeed from the 90 degree or perpendicular $\perp$ visualisation of linear algebra but was by the mathematicians extended broadly to cover such type of relation that corresponds to eigen-properties under given constraints. From this point of view one should say orthogonal is a lent term that was used more and more expanded in its application to genuine independency and not restricted to be set identical to the term perpendicular.

Example: generally applicable you can test any $A$ and $B$ relatively to each other and under your given constraints with respect to orthogonality. A beautiful one might be that two theorems $T_1$ and $T_2$ could be orthogonal.

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The notion of eigenvector does not require, and hence has no relation to, an inner product structure. On the other hand the latter is essential for the definition of orthogonality. – Marc van Leeuwen Jul 14 '13 at 8:35
Marc, this was an answer related intuitive roots of orthogonality, not only what is definition in algebra. The inner product equal zero definition applies of course as definition in algebra, you are right. The question however makes to me to be far more profound and braoder thought. Otherwise one may also expand the classical deifinition of inner product. – al-Hwarizmi Jul 14 '13 at 8:41
I'm having trouble understanding your answer. Could you give some concrete mathematical details? – Jack M Jul 14 '13 at 9:58
As another example you can regards orthogonality in the mathematical context of morphisms e.g. isomorphisms. – al-Hwarizmi Jul 14 '13 at 11:05