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Math Help - abstract vector space

  1. #1
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    abstract vector space

    Hello all- I'm returning to the Maths after a hiatus and I'm already regretting it Very basic question

    determine whether each set with the given operations is a vector space. If not ID the axiom that fails

    for a0 + a1x with the operations

    (a0 + a1x) + (b0 + b1x) = (a0 + b0) + (a1 + b1)x

    and

    k(a0 + a1x) = (ka0) + (ka1)x


    -----------------------------------------------------
    My original thought is that x is a scaler.......but maybe not

    u
    = a0, a1x v = b0,b1x

    u
    + v = v + u

    a0 + b0 = b0 + a0
    a1x + b1x = b1x + a1x = (a1+b1) x

    so closed under addition

    the sencond part looks fine to me so closed under multiplication because the scaler can be applied to each component individually.

    I'm having trouble with the Zero vector (which can be 0 or anything else????) For the zero vector if a0 or a1 are zero than multiplying by any c will give 0 so it passes.
    -----------------------------------------------------------

    The follow up Question True or False

    The set of polynomials with degree exactly 1 is a vector space under the operations defined in exercise 12?

    Back of the book Answer says false????????????? I know I'm blowing it somewhere (probably everywhere) please help nudge me in the right direction.

    Anthony
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  2. #2
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    Re: abstract vector space

    That x is no scaler. So should I not have them in component form of a vector. Still confused
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  3. #3
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    Re: abstract vector space

    your "vectors" are, in this case polynomial functions of degree 1 or less. x is not a scalar, it is what we call an "indetermininate" (which is pretty much a fancy word for "symbolic place-holder"). there is no real difference between:

    a0 + a1x and

    a0 + a1t.

    of course, that's not "entirely" true, because we CAN think of a polynomial as a function (over some number field):

    F--->F

    which, when you input the number x, gives you the number a0 + a1x.

    (in this view we often denote such a polynomial as p(x) = a0 + a1x, although this is slightly incorrect: p(x) is the value of the function at x, the function itself (which is really what our vector is) is just p).

    the a's are "variable constants", they are place-holders for numbers that we may not have explicitly said what they are, but once written down, they're not supposed to change. by contrast the "x" is a "true variable", even after having specified the a's, it is allowed to be any possible element of the domain of p (which is why we call it an "indeterminate").

    it might be more accurate to call "p" the function p = a0 + a1*____, but such a notation is unwieldy and would quickly become confusing in dealing with powers of "___". so we pick a symbol (usually x or t) to indicate what we are going to do with the a's (which are called coefficients).

    i want you to forget anything about what you think vectors are. vectors can be ANYTHING in which we have a notion of "addition" and "scalar multiplication" that follow certain rules. the rules are "the vector space axioms", and they're are like the king's law: their say is final. everything that is true about vector spaces in general is because of one or more of those rules. points in space (position vectors), velocity vectors, force vectors, "arrows with heads and tails" in the plane, real numbers, and many other things are just EXAMPLES of vectors. what a vector IS, is nothing more than ANYTHING which follows the vector space rules. you'll be much better off getting used to this idea now.

    **************

    so...anyway, you are given a set. let's be explicit about what this set is:

    P = {polynomial functions of the form a0 + a1x}, that is to say, polynomial functions of degree 1 or less (and 0: 0 is a "special case" in polynomials, it doesn't really have a degree (is it 0? or 0 + 0x? or 0 + 0x + x2? the reason it doesn't have a degree, is because it can be considered a polynomial of ANY degree, or better yet, of "less than 0 degree").

    now, i hate to be nit-picky, but if you are going to study "abstract" vectors, you must know that EVERY vector space comes with a notion of its "field of scalars". you don't have a vector space until you say what "scalars" are. fields are another abstract entity, which have THEIR OWN set of axioms to follow (so if you think about it, there's "hidden vector space axioms" (the field axioms) we don't talk about very much).

    so, we can't really talk about polynomial functions, until we say "WHERE the coefficients" come from. for most of the time, the scalars will be one of three types of numbers:

    1) rational numbers
    2) real numbers
    3) complex numbers

    you don't say what your field of scalars is, so i will assume it is the field of real numbers (that's the one used most commonly). so we can restate our set as:

    P = {polynomial functions from R to R of the form: p(x) = a0 + a1x : a0, a1 in R}

    do we have an addition defined on P? why, yes, we do, it is given by the formula:

    p(x) + q(x) = (a0 + a1x) + (b0 + b1x) = (a0+b0) + (a1+b1)x

    for example:

    (2 + 3x) + (-7 + x) = (2 - 7) + (3 + 1)x = -5 + 4x

    do we have a scalar multiplication on P (this is just a function FxV--->V, you feed it a scalar on the left, a vector on the right, and it spits out some other vector)?

    again, yes we do, the formula we are given is:

    k(p(x)) = k(a0 + a1x) = (ka0) + (ka1)x.

    for example:

    (1/2)(2 + 3x) = (1/2)(2) + ((1/2)(3))x = 1 + (3/2)x.

    now, you have to verify that these two operations obey all the axioms. you probably have about 10 of them, in all.

    a word about the 0-vector in a vector space. DO NOT get this confused with the "scalar 0" (often both are simply denoted by "0"), these are TWO different things. the 0-vector is a vector z such that:

    v + z = z + w = v for ALL vectors v, that is it is a "vector addition identity". the property i have just listed DEFINES it. in some vector spaces the 0-vector may not look very "zero-ish". that doesn't matter. if it is a vector z such that:

    v + z = z + v = v, for any vector v, it's the 0-vector. that's how you check. the property c(z) = z does NOT define the 0-vector z, it s a property of the 0-vector that is DERIVED from the vector space axioms:

    proof: (get used to these)

    let z be the 0-vector, and let c be any scalar. then cz = z.

    by (some other axiom...usually called a "distributive law" although again that's not strictly accurate):

    c(u + v) = cu + cv, for all vectors u,v and all scalars c. in particular, if we take u = v = z:

    c(z + z) = cz + cz

    but since z is the 0-vector, z + z = z. therefore:

    cz = c(z+z) = cz + cz.

    thus:

    cz + (-cz) = (cz + cz) + (-cz) = (cz + (cz + (-cz)) <---by the associative law

    z = cz + z (by the "additive inverse law" that says v + -v = z, for all vectors v)

    z = cz (because z is the additive identity, so cz + z = cz).

    i deliberately used "z" instead of "0", to get you in the habit of treating the 0-vector and the 0-scalar differently. they are different things, they obery different rules (axioms).

    coming back to our set P, one of the vector space axioms says that "there exists" a 0-vector. so we need to FIND one in P.

    luckily, it's not hard to find, as you yourself pointed out the function:

    0(x) = 0 + 0x (picking a0 = a1 = 0) will work nicely: for ALL p in P-

    p(x) + 0(x) = (a0 + a1x) + (0 + 0x) = (a0 + 0) + (a1 + 0)x (the 0's in the parentheses are now "scalar 0's")

    = a0 + a1x = p(x).

    if you have already established that p(x) + q(x) = q(x) + p(x), for all p, and q in P, then we don't need to prove:

    0(x) + p(x) = p(x).

    ******************************

    for the follow up question, i give you another question:

    if p(x) = 1 - x, and q(x) = x, is p(x) + q(x) a polynomial of degree exactly 1?
    Last edited by Deveno; January 23rd 2013 at 08:31 PM.
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  4. #4
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    Re: abstract vector space

    (a0 + a1x) + (b0 + b1x) = (a0 + b0) + (a1 + b1)x
    and
    k(a0 + a1x) = (ka0) + (ka1)x

    Above satisfies requirement of a vector space V: If a and b belong to V then so do a + b and ka, k a scalar (real number for now).

    A polynomial of degree exactly 1 does not satisfy requirement because it excludes 0, which is not degree 1.
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  5. #5
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    Re: abstract vector space

    Quote Originally Posted by Hartlw View Post
    (a0 + a1x) + (b0 + b1x) = (a0 + b0) + (a1 + b1)x
    and
    k(a0 + a1x) = (ka0) + (ka1)x

    Above satisfies requirement of a vector space V: If a and b belong to V then so do a + b and ka, k a scalar (real number for now).

    A polynomial of degree exactly 1 does not satisfy requirement because it excludes 0, which is not degree 1.
    these are NOT the requirements of a vector space. those are the conditions that a subset of a GIVEN vector space be a SUBSPACE. we are not explicitly given a "parent space" so ALL the vector space axioms must be verified.
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  6. #6
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    Re: abstract vector space

    a+bx <-> (a,b)
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  7. #7
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    Re: abstract vector space

    which means, what, exactly? i know there is a linear isomorphism between his P, and R2, but at this stage of the game, it is unlikely his text has even defined what a linear isomorphism IS.
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  8. #8
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    Re: abstract vector space

    Wow thanks. if p(x) = 1 - x and q(x) = x; p(x) + q(x) = (1-x) + x = 1. therefore it's no longer a polynomial of degree 1????????????? yes no ? By the way all the definitions and syntax is what's going to kill me. Do you have any suggestions for references that encompass a fair amount of general info for the removed student. I'm going through Diffe Q and pursuing a ChemE major
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  9. #9
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    Re: abstract vector space

    yes, constant polynomial are of degree 0 < 1.

    here's the thing. linear algebra is a powerful tool for solving differential equations. but to "use it right" you have to know what you are doing, so you don't suffer from the "garbage in, garbage out" syndrome. some of the things that matter in differential equations:

    1) vectors can be functions (if you think about this, it makes sense: we can add functions together, and "scale them" by multiplying them by a constant (scalar)).

    2) we can have linear functions OF functions. the best example of this is the derivative: D(f+g) = Df + Dg, and D(af) = aDf (here, "Df" is another way of writing d/dx(f(x)), the derivative of f w.r.t. x).

    just like in "high-school algebra" there are rules that allow us to simplify calculations. these are mainly the same rules as we're used to, but:

    a) vectors are not numbers (sometimes it's best to to think too hard on "what they are", this limits our intuition)

    b) NO DIVISION!!!!!!

    c) "multiplication" does not act like you are used to...there are several "kinds of products"...scalar products, inner products, cross-products, and matrix products. none of these behave like ordinary multiplication (the rules are different).

    so when you see something like "av" you have to stop and think: "what is a, what is v, and where does av live".

    there used to be a good online reference called "Paul's Notes" which he has taken down. in its stead, you might want to check out the videos at https://www.khanacademy.org/
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