
#1
Dec512, 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 ndimensional 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 n1 degrees of freedom? If so, why? Examples To make my question more clear, here are two instances of this scenario.




#2
Dec512, 12:57 AM

P: 2,475

this kind of reminds me of a holographic universe where you have 3 dimensions but can map everything to 2.




#3
Dec512, 01:42 AM

P: 166

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



#4
Dec512, 07:34 AM

Math
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P: 38,879

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 n1 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.




#5
Dec512, 10:17 AM

Engineering
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P: 6,344

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". 



#6
Dec512, 06:37 PM

P: 46

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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