Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks.

share|improve this question
    
I'm not sure what your exact question is, but this may help: en.wikipedia.org/wiki/Dot_product#Geometric_interpretation –  wj32 Oct 19 '12 at 0:45

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.