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There's something that's always bothered me in my encounters with linear algebra, it seems like every vector they ever give is already defined according to some basis: e.g. A=<2,-1,0,5,6>.

Imagine that I have a very large circular sheet of paper so that there is no obvious x and y (or i and j). I can put a dot somewhere on the paper and draw an arrow and call it V. Then, from the same dot, I can draw another arrow and call it e1 (my first basis vector). Now, without going into details, if I just have a piece of string and a straight-edge (no rulers, protractors or other pre-made measuring devices) I can measure the length of V in terms of the length of e1 (which I take to be 1-unit long), and I can use the string to make circles and I can use the string to measure arc-lengths in terms of e1 and I can figure out the angle (in radians) between V and e1. And, with just the string and straight-edge, I can even construct e2 perpendicular to e1.

I can figure out how to do all of this in 2-dimensions without relying on a pre-existing set of basis vectors, but it requires me to actually physically draw and measure things.

My question is how do you do this in n-dimensions? Suppose I have a vector in 4-dimensional space, but that space does not yet have any defined basis. So, I define a second vector as e1. To the best of my knowledge 4D paper and 4D string do not exist in the physical world, so I have to do it abstractly. How do I determine the relative length of my two vectors and the angle between them?


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I'm not sure what your exact question is, but this may help: – wj32 Oct 19 '12 at 0:45

Well, you can take the vector space made up of continuous real-valued functions on the closed interval $[0,1],$ then say that the length of any single vector $f(x)$ is defined to be $$ |f| \; = \; \sqrt{\int_0^1 \; f^2(x) \, dx \;} $$ and the inner product between two vectors $f,g$ is $$ \langle f,g \rangle \; = \; \int_0^1 \; f(x) g(x) \, dx \; $$ and it all works, absolutely everything works, with no evident basis.

One could also point out that the dot product, working as it does in $\mathbb R^n,$ is a theorem by induction on the (finite) dimension, with the basic fact being the Pythagorean Theorem.

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Thank you, Will. But maybe there is something that I'm missing. When I looked at your explanation I decided to test it out on the Taylor series for e^x = 1 + x + x^2/2 + x^3/3! ... I figured if I used your method with f(x)=e^x and g(x) = x^2, I should end up with <f,g>/|g| = 0.5. That is: e^x "points" 1 in the x^0 direction, 1 in the x^1 direction, 0.5 in the x^2 direction... but instead of 0.5, I come up with 1.606. So, I guess I'm missing something... – user45177 Oct 19 '12 at 1:40
@user45177, you are incorrect in assuming that the $x^n$ make an "orthonormal basis." This is not the case. $$ \langle x^m, x^n \rangle \; = \; \frac{1}{1 + m + n}. $$ The Taylor series for $e^x$ does not tell you anything useful about the inner product of $e^x$ and some fixed $x^m.$ – Will Jagy Oct 19 '12 at 1:47

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