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Prove 1x = x from linear space axioms 
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#1
Dec1009, 02:57 PM

P: 76

1. The problem statement, all variables and given/known data
"Prove that Axiom 10[, the existence of identity in a linear space,] can be deduced from the other axioms." 2. Relevant equations Now, I know that these axiom numberings are fairly arbitrary, but I'll put them in anyways for easy reference. I'm listing them word for word in case there's a subtlety I'm missing. "Let V denote a nonempty set of objects, called elements. The set V is called a linear space if it satisfies the following ten axioms...: Axiom 1. Closure Under Addition. For every pair of elements x and y in V there corresponds a unique element in V called the sum of x and y, denoted by x + y. Axiom 2. Closure Under Multiplication by Real Numbers. Axiom 3. Commutative Law. For all x and y in V, we have x + y = y + x. Axiom 4. Associative Law. For all x, y, and z in V, we have (x + y) + z = x + (y + z). Axiom 5. Existence of Zero Element. There is an element in V, denoted by O, such that x+ O = x for all x in V. Axiom 6. Existence of Negatives. For every x in V, the element (1)x has the property x + (1)x = O Axiom 7. Associative Law. For every x in V and all real numbers a and b, we have a(bx) = (ab)x. Axiom 8. Distributive Law for Addition in V. For all x and y in V and all real a, we have a(x + y) = ax + ay Axiom 9. Distributive Law for Addition of Numbers. For all x in V and all real a and b, we have (a + b)x = ax + bx. Axiom 10. Existence of Identity. For every x in V, we have 1x = x." Afterwards, it states that when the scalars are real, such a linear space is sometimes called a real linear space while if the scalars are complex numbers, then the space is called a complex linear space. It also states that linear spaces are sometimes reffered to as linear vector spaces or vector spaces. The book states that whenever it refers to a linear space without designation, the space could be either real or complex. Also, the uniqueness of the zero element and negative elements are proved as theorems later on. And while the book never explicitly says so, from sample proofs of theorems from the axioms I can see that substitution is another, though "hidden," assumption. 3. The attempt at a solution I am quite stuck on this one. It's an axiom so shouldn't it be impossible to prove it from the other axioms? I think I can prove it if I assume the standard rules of multiplication of real numbers. They are real scalars after all, so shouldn't that be fine (after I do the same assuming the scalars are complex)? However, I have a problem with that since the statement explicitly said to prove it from the other axioms of a linear space, so I would assume that's all I can use. Anyways, this is as far as I've gotten: By Axiom 2, there is an element in V, say z, such that z = 1x. By Axiom 6, z + (1)z = O. Thus, 1x +(1)1x = O. By Axioms 9 and 6: 1x +(1)1x = (1 + (1)1)x = Ox. Therefore, 1x + (1)1x = Ox = O = x + (1)x (last part by Axiom 5). Now, if I could somehow prove 1 + 1 = O, 1x + (1)x = O, or x + (1)1x = O, then I could finish the proof by adding x (or 1x) to both sides of 1x + (1)1x = x + (1)x getting 1x + (1)1x + x = x +(1)x + x = 1x + O = x + O = 1x = x. (1 + 1 = O would allow me to show 1x + (1)x = Ox = O). Of course, one problem is the lack of a unique zero element, but I suppose I would have to prove that too along the course of my proof? Or do I need to stick strictly to the Axioms? I'm probably missing something simple here, if this can be proved at all. 


#2
Dec1009, 03:29 PM

P: 424

You have already worked out: 1x + (1)1x = x + (1)x Now by associativity: (1)1x = (1*1)x = (1)x so: 1x + (1)x = x + (1)x By the existence of negatives (1)x has a negative y which we can add to get: 1x = 1x+ O = 1x+(1)x + y = x + (1)x + y = x + O = x 


#3
Dec1009, 04:37 PM

P: 76

I can see why 1 + 1 =/= O. It's because O for the reals is 0 (a scalar, like you said). Thus I should have had 0x = O. Correct?
And, I could prove it that way, and it was that which I had originally had in mind for proving it. However my problem was whether I could assume the identity axiom of real numbers which 1*1 = 1 is doing (since you obviously can't assume the identity axiom of linear spaces for this). I understand that the set of real numbers is an example of a linear space, but then isn't assuming 1*1 = 1 assuming the identity axiom of linear spaces (with x = 1) and so is a subtle form of circular logic? Or maybe it's that the set of real numbers are assumed to exist since the scalars are assumed to be real in the axioms and thus any operations with real scalars must satisfy the axioms of real numbers? Also, afterwards I'm supposed to prove that if Axiom 6 is replaced by Axiom 6' (For every x in V there is an element y in V such that x + y = O.) then 1x = x cannot be deduced from the other axioms. I'm not sure how I would prove such a thing. I thought a proof by contradiction would work by assuming 1x = x could be deduced from the axioms, but I'm not sure where to start after that (or really, what contradiction I'd be looking for). Could I have some help on this too? 


#4
Dec1009, 05:00 PM

P: 424

Prove 1x = x from linear space axioms
The real numbers is a linear space, but we don't build it on the theory of linear spaces. Instead assume all axioms except axiom 6, and also assume axiom 6'. Given some vector x you want to show x+(1)x =0. According to axiom 6' you have some vector y such that x+y =0 so it would suffice to show y=(1)x. 0= 0x = (1 + (1))x = 1x+(1)x = x + (1)x = x + y Cancel x by adding y again and then you have axiom 6. 


#5
Dec1009, 05:17 PM

P: 76

Thus, multiplication (and addition) of scalars retains all the properties of the particular special case of real numbers. All of these properties of this special case can be used to prove statements about the general object and about other special cases of the general object. So the special case of real numbers then is independent of the general linear space since the special case of real numbers was defined axiomatically and independent of the axioms for the general linear space (instead of being defined, as you put it, "real numbers are those objects which satisfy axioms 19, implying 10 too, under certain rules of addition and multplication," which would be circular logic), which means real numbers exist independently of the general linear space? I must admit I'm still having a tough time wrapping my head around the logic. If that's the case, will I also need to prove this assuming the scalars are complex? Edit: Oh! Is this a way to think about it? A special case of a fruit is an apple. A property (by an axiom) of apples is that 1*apple = one apple. A property to prove for a fruit is that 1*fruit = one fruit. It happens that apples are related to fruit by multiplication of scalars and addition. The property 1*apple = one apple can be used to prove in general 1*fruit = one fruit when used appropriately as apples are related to fruit and in conjunction with other axioms for fruit. 1*apple = one apple as an intrinsic property of apples (by an axiom), not by an axiom for fruit, so there is no circular logic. So it just so happens that real numbers are a linear space because they satisfy the axioms for it. That they happen to be used by the general case for finding other special cases is irrelevant. I guess then that scalars can only be either real numbers or complex numbers? And I guess this also means that you don't derive real numbers, complex numbers, realvalued functions, etc. from linear spaces since they have their own independent existence, but you can derive others which do depend on the linear space? Edit2: Would it be true that if an Axiom could be deduced from other Axioms, it is really a theorem and so is redundant to the definition of a linear space? A special case then would need only sastisfy the other Axioms. If it could be shown that there was a special case (of this linear space without the redundant axiom) did not satisfy the redundant axiom, then it would show that the redundant axiom is not redundant and could not be deduced from the others because any object satisfying those axioms would have to satisfy the redundant axiom (the "theorem") because it was deduced from the other axioms and thus should be true for all objects satisfying those axioms. Thus, if I were to assume Axiom 10 could be deduced from Axioms 15, 6', and 79 then all I would need to do would be to find an object satisfying 15, 6', and 79 but not satisfying Axiom 10? This would arrive at a contradiction which would show Axiom 10 could not be deduced from those Axioms alone which would prove the statement. Is this right? 


#6
Dec1009, 05:54 PM

P: 424

For simplicity let us simply try to work on the real numbers as a real linear space with the ordinary addition, but with a new type of scalar multiplication. An idea that immediately springs to mind is to define scalar multiplication in such a way that there are some element y of the linear space which cannot be expressed in the form rx, then we can't have 1y=y. We know that 0x = 0 so why not take it to extremes and let 0 be the only element scalar multiplication can give; so define rx = 0 for all scalars r and vectors x. Verify that all axioms except 6 and 10 are satisfied. 


#7
Dec1009, 06:15 PM

P: 76

Axiom 6' is satisfied since there exists a real number y and real number x such that x + y = 0 by properties of real numbers. Axiom 7 is satisfied because a(rx) = a0 = 0 = (ar)x = 0x the definition of this kind of scalar multiplication. Axiom 8 is satisfied since r(x + y) = 0 = 0 + 0 = rx + ry. Axiom 9 is satisfied (letting z = r + a, where z is a real number, which is justified by Axiom 1) since (r + a)x = zx = 0 = 0 + 0 = rx + ax. However, Axiom 10 is not satisfied because 1x = 0 =/= x by this definition of scalar multiplication. Thus, Axiom 10 cannot be deduced from the other Axioms given Axiom 6' is substituted for Axiom 6. I'm not sure why I'd need to show this kind of object did not satisfy Axiom 6 (will you please let me know why?). But here is why anyways: x + (1)x = x + 0 =/= 0, so it does not satisfy Axiom 6. Thanks a lot for the help! 


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