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In the space $\mathbb{R}^n$, by definition, the component representation of $e_1, e_2, ..., e_n$ with respect to basis $\{ e_1, e_2, ..., e_n \}$ is (1, 0, ..., 0), (0, 1, 0, ..., 0), ..., (0, 0, ..., 1).

Without equipping this space with a choice of inner product, it is not defined whether any pair of these basis vectors are orthogonal, nor is their length defined.

Yet it is a common habit to visualize the $\mathbb{R}^3$ vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1) along a left-right horizontal axis, a front-back horizontal axis, and a vertical axis, and furthermore imagine the tick at '1' to be the same distance from the origin along all of these axes.

Is the act of visualizing these this way somehow inducing the choice of the dot product as the inner product? Or, is there any way to rationalize and connect the visual imagination being performed here with the mathematical assumptions it induces?

Henry Bigelow
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  • It would be (arguably) presumptuous to depict basis vectors as orthogonal without any inner product already there, but $\Bbb R^n$ already does have an inner product so there's no real question here. – anon Apr 08 '18 at 19:21
  • A way to remember the formula for the linear correlation co-efficient $L$ of two sets of data $(x_1,...,x_n), (y_1,.,y_n)$ with $n>1:$ Let $x'_i=x_i-M_x$ and $y'_i=y_i-M_i,$ where $M_x$ is the mean of $x_1,..,x_n$ and $M_y$ is the mean of $y_1,...,y_n .$ Let $X'=(x'_1,...,x'_n)$ and $Y'=(y'_1,...,y_n).$ Let $O$ be the origin in $\Bbb R^n.$ If $X'\ne O\ne Y$' then $L=\cos \angle X'OY'= X'\cdot Y'/|X'|\cdot |Y'| .$. – DanielWainfleet Apr 08 '18 at 21:25
  • The geometric significance of the inner product is, I think, a fairly recent discovery. It allows us to discuss the geometry of an $m$-dimensional hyper-plane $P$ in $\Bbb R^n$ (with $m<n$) without the need for a different co-ordinate system for $P. $ For example the formula for $\cos \angle X'OY'$ in my prior comment . – DanielWainfleet Apr 08 '18 at 21:36
  • In my first comment there is a typo, $M_i$ should be $M_y$. – DanielWainfleet Apr 08 '18 at 21:45

2 Answers2

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Is the act of visualizing these this way somehow inducing the choice of the dot product as the inner product?

Yes. The general inner product of the space $\mathbb{R}^n$ (considered to be column vectors) is $$ <x, y>= x^Tgy $$ where $g$ is some symmetric matrix. For example, in $\mathbb{R}^2$ $$ <x, y>= a x_1 y_1 + b x_2 y_2 + c (x_1 y_2 + x_2 y_1). $$

Thus the basis vectors $(1,0)$ and $(0,1)$ may not be unitary or orthogonal. The matrix $g$ can be considered to contain, as elements, the dot products of vectors, and is known in differential geometry as the metric tensor. To be a true inner product we require $g$ to be positive definite (so that the magnitude of all non-zero vectors is positive), but in differential geometry this requirement is often dropped.

Eddy
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  • Thanks Eddy. Just a quick follow up question: Supposing I instead choose a = 1, b = 1, and c = 0.99 for my g matrix, then I should be visualizing my basis vectors as unitary, but almost parallel? – Henry Bigelow Apr 09 '18 at 20:56
  • If you want to figure out how a weird inner product space could be constructed from a dot product (which us what I think you're asking) then we need to make the interpretation that the components of the vectors are the coefficients of some basis vectors in an 'equivalent' dot product space. That is if $x = (x_1,\ldots,x_n)$ then we identify this vector with the vector $X=x_1 e_1 +... + x_n e_n$ where $e_1,...,e_n$ are n-tuple basis vectors in the new space, defined by $X \cdot Y = <x, y>$. From this we can identify $(1,0) $ and $(0,1) $ in your example with appropriate vectors in the... – Eddy Apr 09 '18 at 21:08
  • ... dot product space. – Eddy Apr 09 '18 at 21:09
  • Reading what I've written it isn't that clear, let me know if you'ld like me to expand tge answer to include interpretation. – Eddy Apr 09 '18 at 21:15
  • Hi Eddy, your original answer led me down a productive path, and I consulted the Georgi Shilov text, chapters 7 and 8. I still have some confusion on this issue though, but I can at least write a more clear question. Why don't I do that, and then link to the question here? – Henry Bigelow Apr 09 '18 at 21:47
  • Go ahead aaaaaa – Eddy Apr 09 '18 at 21:48
  • Hi Eddy, just a quick follow up. I tried drawing a unit circle in R^2, using basis vectors e1 (horizontal and to the right), and e2 (pointing just slightly sloping upward from e1). The inner product matrix for this would be (1, 0.9, 0.9, 1) or so. To my surprise, the unit circle does indeed look like a circle! As the off-diagonal elements of the inner product matrix approach 1 (and thus e2 approaches e1), the components of a vector at the apex of the circle approach (-inf, inf), yet its length is still 1, both as calculated and as graphed! I'm not sure how to interpret this just yet. – Henry Bigelow Apr 10 '18 at 22:48
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As you noted, the vectors of any basis in $\mathbb{R}^3$, represented in the same ''themself'' basis, have components $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$.

More, for any basis ve can define an inner product in $\mathbb{R}^3$ such that the vector of the basis are orthogonal ( see here).

Usually we represents the vectors $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$ as a triple of orthogonal vector like the three sides of a cube, but this is only a consequence of a bias deriving from our physical experience, because in our world there is a ''vertical'' direction and a ''horizontal'' plane (the horizon) determined by gravity.

I don't see a pure mathematical reason to represents ''orthogonality'' only in this usual way..... But maybe I'm wrong. I have a question about this but without an answer.

Emilio Novati
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