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Homework Help: Prove 1x = x from linear space axioms

  1. Dec 10, 2009 #1
    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.
    Last edited: Dec 10, 2009
  2. jcsd
  3. Dec 10, 2009 #2
    This doesn't make any sense as O is a vector, not a scalar. It's not true Ox = O as there is no general multiplication of vectors. What you want is as you noted 0x = O.

    You have already worked out:
    1x + (-1)1x = x + (-1)x
    Now by associativity:
    (-1)1x = (-1*1)x = (-1)x
    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
  4. Dec 10, 2009 #3
    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?
    Last edited: Dec 10, 2009
  5. Dec 10, 2009 #4
    Yes you should have said 0x = O, and it's true that in the special case of the real number O and 0 coincide, but it's not true in general so for general elements you should not write 1 + (-1) = O. In the same way in the real numbers we can perform multiplication, but this is not true in an arbitrary linear space so you can't use multiplication of vectors.

    No because earlier you have showed/defined 1 to have this behavior. You don't know whether a general linear space has a scalar identity, but you know this to be true in the specific case of the real numbers. If you look back to the book that proved that the real numbers had a multiplicative identity 1 and it simply stated: the real numbers satisfy the axiom 1-9 so axiom 10 is satisfied, then you would have circular logic, but they probably performed an explicit argument.

    The real numbers is a linear space, but we don't build it on the theory of linear spaces.

    I don't really think contradiction is a good choice here since you're supposed to prove that something is true, but its negation is not simpler.

    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.
  6. Dec 10, 2009 #5
    Okay, thanks for that!

    Let me see if I understand now then. We know that for a special case of a general object, the identity axiom holds (my book didn't prove 1x = x for the real numbers but listed it as an axiom for real numbers waaaay earlier on). It just so happens that the way this general object is defined, this particular special case is related to the general object through multiplication of scalars (and addition in the case of the Distributive Axiom of the general linear space).

    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 1-9, 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, real-valued functions, etc. from linear spaces since they have their own independent existence, but you can derive others which do depend on the linear space?

    Oops! Sorry about that. I made a typo. I edited the question to what it was supposed to be which was "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 Axiom 10 cannot be deduced from the other axioms."

    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 1-5, 6', and 7-9 then all I would need to do would be to find an object satisfying 1-5, 6', and 7-9 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?
    Last edited: Dec 10, 2009
  7. Dec 10, 2009 #6
    Well the definition you gave for a linear space is really just a real linear space (since you explicitly call the scalars real in several axioms), but if you also consider complex vector spaces, then yes, but you don't really need to redo your work just note that 1+(-1)=0 also holds in the complex numbers since the real numbers are embedded in the complex numbers. In fact we usually define -x in more general contexts to mean exactly the element such that x + (-x) = O, so everything would also works if you later extend the concept of linear space to be over more general scalars (you will probably do this later in your course and extend the scalars to all kinds of fields).

    Well in that case contradiction is a good idea. We should construct a space which satisfy axiom 1-5,6', and 7-9, but not 10.

    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.
  8. Dec 10, 2009 #7
    Okay, I'm understanding now. That sounds soooo cool! Anyways, thanks for that. It answered one of the questions in one of my edits too!

    Okay, that answers the question of another edit! So then, all the addition Axioms are satisfied since addition is still defined the same way. Axiom 2 is satisfied since for every element x in V and real scalar r, the product rx exists by properties of real numbers. Axiom 5 is satisfied since x + rx = x + 0 = x since this is real number addition defined in the normal manner.

    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!
    Last edited: Dec 10, 2009
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