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Question about counting degrees of freedom

by samh
Tags: counting, degrees, freedom
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samh
#1
Dec5-12, 12:52 AM
P: 46
Suppose that for some application it is mathematically convenient to represent certain objects of interest (e.g., lines or conics) as n-dimensional vectors. That such a representation exists lets us conclude that in order to specify such an object, no more than n values are necessary. That is, there are at most n degrees of freedom.

But now suppose I tell you that this representation has the following special property: any two vectors that are scalar multiples of one another represent the same object (provided that the scale factor is nonzero). So here’s my question: does this allow us to conclude that there are, in fact, at most n-1 degrees of freedom? If so, why?

Examples
To make my question more clear, here are two instances of this scenario.
  1. Lines in the plane, which have 2 DOF, can be expressed in the form ax+by+c=0, and hence we may choose to represent them as 3-vectors (a,b,c). Since (ka)x+(kb)y+(kc)=0 for nonzero k represents the same line, (a,b,c) is equivalent to all scalar multiples (ka,kb,kc).
  2. Transformations that act on homogeneous coordinates are generally defined only up to scale. For example, 3x3 homographies have 8 DOF rather than 9 "because" they are defined only up to scale.
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jedishrfu
#2
Dec5-12, 12:57 AM
P: 2,805
this kind of reminds me of a holographic universe where you have 3 dimensions but can map everything to 2.
justsomeguy
#3
Dec5-12, 01:42 AM
P: 166
does this allow us to conclude that there are, in fact, at most n-1 degrees of freedom? If so, why?
I don't think it does. If it did, then points on a 1D line would have 0 degrees of freedom, which is clearly a contradiction, as there are infinitely many primes that you can place on the line.

Of course, I'm not even an armchair mathematician.

HallsofIvy
#4
Dec5-12, 07:34 AM
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Question about counting degrees of freedom

You could, for example, divide the entire vector by, say, the first component and so represent any vector by <1, ....> with n-1 arbitrary numbers. To answer justsomeguys objection, if n were 1, every point would be represented by a single number but, since multiplying any such by a number, they all represent the same vector. The space containing a single vector does, in fact, have dimension 1. I have no idea what "primes" have to do with this.
AlephZero
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Dec5-12, 10:17 AM
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Quote Quote by HallsofIvy View Post
You could, for example, divide the entire vector by, say, the first component and so represent any vector by <1, ....> with n-1 arbitrary numbers.
.... unless the first component was zero!!!

It is often more convenient to use more "values" than are strictly necessary to describe the degrees of freedom of the system, because it avoids a lot of special cases. For example you could write the equation of the line as y = mx + c with only two constants, but then you can't deal with vertical lines unless you add the (large) complication that m can equal "infinity".
samh
#6
Dec5-12, 06:37 PM
P: 46
Quote Quote by AlephZero View Post
.... unless the first component was zero!!!
This makes sense. Then it makes me wonder, since you can't assume WLOG that the first component is 1, how do we argue that there are n-1 DOF rather than n?

What about this simpler case: we want to solve for [itex]x\in\mathbb{R}^9[/itex] and to do so we collect constraints [itex]a_i^Tx=b_i[/itex]. If we can find 9 of these, we can solve for [itex]x[/itex]. But if all [itex]b_i[/itex] are zero (and we want [itex]x\neq0[/itex]) we need only collect 8 constraints! Why exactly is this? How would you prove it?


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