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Vector space several questions

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    (i) Is the set of all mappings f:R3->R

    a)f(x,y,z)=ax+by+cz

    b)f(x,y,z)=ax+by+cz+d

    for a,b,c,d Є R, vector space, for the standard addition operation and scalar multiplication of function with real number?

    (ii) Is the set of all vectors in R3 which are collinear with the vector (1,-2,3), vector space for the standard addition operation in R3 and multiply of vector from R3 with scalar?

    (iii) Is the set of all vectors x(2,0,3) + y(-1,4,-2) where x,y Є R, vector space for the operations in R3(R)?

    2. Relevant equations

    [f+(g+h)](x)=f(x)+(g+h)(x)=f(x)+g(x)+h(x)=(f(x)+g(x))+h(x)=[f+(g+h)](x)

    (f+g)(x)=f(x)+g(x)

    (kf)(x)=kf(x)

    (u+v)+w=u+(v+w)

    (ab)u=a(bu)

    3. The attempt at a solution

    (i) Should I check if it's vector space by using:

    f(x,y,z)=ax+by+cz

    (f+g)(x)=f(x)+g(x) and (kf)(x)=kf(x), but what will be g(x) ?

    (ii) If it is collinear with the vector (1,-2,3), probably the set of vectors lies on same line.
    So the set of vectors are λ(1,-2,3).

    u=a(1,-2,3)
    v=b(1,-2,3)

    u+v=a(1,-2,3)+b(1,-2,3)=(a+b,-2a-2b,3a+3b)=[a+b,-2(a+b),3(a+b)]

    Really, I don't know what is the proof here.

    Please help.
     
  2. jcsd
  3. Jun 16, 2009 #2

    CompuChip

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    In the first one, you can take two arbitrary functions
    f(x, y, z) = ax + by + cz
    g(x, y, z) = a'x + b'y + c'z

    In general, can you state the axioms of a vector space (i.e. what you have to check)?
     
  4. Jun 16, 2009 #3
    I think I need to check:
    (f+g)(x)=f(x)+g(x)

    (kf)(x)=kf(x)

    But how will I do that?
     
  5. Jun 16, 2009 #4

    CompuChip

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    No, that is the definition of (f + g) and (kf).

    As I asked before: what are the defining properties of a vector space?
     
  6. Jun 16, 2009 #5
    Sorry for misunderstanding.

    Standard vector addition and scalar multiplication are:

    [tex](x_1,x_2,...x_n)+(y_1,y_2,...,y_n)=(x_1+y_1,x_2+y_2,...,x_n+y_n)[/tex]

    and

    [tex]k(a_1,a_2,...,a_n)=(ka_1,ka_2,...,ka_n)[/tex]

    In this case we got R3, therefore [itex](x_1,x_2,x_3)+(y_1,y_2,y_3)+(z_1,z_2,z_3)=(x_1+y_1,x_2+y_2,y_3+z_3)[/itex]

    With other words u+v is addition and cu is scalar multiplication.

    Probably I need to check them.

    Or I need to check all of the axioms from (a) to (j)

    Here is the picture of all axioms:

    http://img196.imageshack.us/img196/7121/79714632.th.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  7. Jun 16, 2009 #6

    Mark44

    Staff: Mentor

    No, just a sum of two vectors is sufficient.
    [itex](x_1,x_2,x_3)+(y_1,y_2,y_3)=(x_1+y_1,x_2+y_2,x_3+y_3)[/itex]

    Yes, to show that some space is a vector space, you need to check all 10 axioms.
     
    Last edited by a moderator: May 4, 2017
  8. Jun 16, 2009 #7
    Thanks for the replies.

    And what will be the vectors, u, v and w?

    I think in my task says that I should prove the axioms using functions:

    C1: [f+(g+h)](x)=f(x)+(g+h)(x)=f(x)+g(x)+h(x)=(f(x)+g(x))+h(x)=[f+(g+h)](x)

    C2: c(x)=0;
    (f+c)(x)=f(x)+c(x)=f(x)+0=f(x)

    C3:
    (-f)(x)=-f(x)
    [f+(-f)](x)=f(x)+(-f)(x)=f(x)-f(x)=0

    C4:
    f(x)+g(x)=g(x)+f(x)

    M1:
    a(f(x)+g(x))=af(x)+ag(x)

    and so on..
     
  9. Jun 16, 2009 #8

    HallsofIvy

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    You should be able to use the general theorem that addition of functions, (f+g)(x)= f(x)+ g(x), is associative and commutative as well as the facts that multiplication of a function by a number, (af)(x)= af(x), is associative and distributes over addition.

    Generally speaking, unless you are given unusual definitions for the operations, you only need to show that the set is closed under addition and multiplication by a scalar to show that a set, together with given operations, is a vector space.

    For part (a), you need to look at f(x)= ax+ by+ cz and g(x)= ux+ vy+ wz and their sum f+ g as well as rf for a number r.
     
  10. Jun 16, 2009 #9
    Thanks for the reply.

    Do you mean f(x,y,z)=ax+by+cz and g(x,y,z)=ux+vy+wz ? But I can't prove anything, because (f+g)(x,y,z)=f(x,y,z)+g(x,y,z)=x(a+u)+y(b+v)+z(c+w)

    f(x,y,z)+g(x,y,z)=ax+by+cz+ux+vy+wz=x(a+u)+y(b+v)+z(c+w)
    g(x,y,z)+f(x,y,z)=ux+vy+wz+ax+by+cz=x(a+u)+y(b+v)+z(c+w)

    and f(x)+g(x)=g(x)+f(x).

    Also there is nothing to prove with (af)(x,y,z)=af(x,y,z)

    Probably, I need to go through all of the axioms.

    But what about (ii) and (iii) ?
     
  11. Jun 16, 2009 #10

    Mark44

    Staff: Mentor

    For (a) you're supposed to prove that the set of mappings f:R3 -> R, defined by f(x, y, z) = ax + by + cz is a vector space. In other words, the the space of function that are defined this way is a vector space.

    Take two functions that belong to this space, f and g. From the given information, f(x, y, z) = ax + by + cz, and g(x, y, z) = ax + by + cz. Now, can you show that f+g belongs to this same space; i.e., that the space is closed under vector addition? You'll also need to show that the space is closed under scalar multiplication; i.e., that cf is also in this space.
     
  12. Jun 17, 2009 #11

    CompuChip

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    Now you are just taking one function. You probably mean g(x, y, z) = u x + v y + w z?

    Yes, for example.

    Yes, that's the definition of the function (f+g)

    Yes. So is this again of the form (f+g)(x, y, z) = px+qy+rz for real numbers p, q, r?

    There is, but it's just as easy.

    You should check all the axioms, although - as said - a few of them you don't usually go through at great length, such as distributivity and associativity of addiiton. But look at (e) for example. There should be a function N(x, y, z) such that for any f(x, y, z) = a x + b y + c z, (N + f)(x, y, z) = f(x, y, z). That is, N(x, y, z) is the special element 0 from axiom (e). First you need to think of such a function, then show that it is actually in the vector space (i.e. you can write it as ux+vy+wz).
     
  13. Jun 17, 2009 #12
    Thanks for the reply.

    Is it something like this?

    f(x)+g(x)=ax + by + cz+ ax + by + cz=2ax+2by+2cz=2(ax+by+cz)=2*(f(x)+g(x))

    cf(x)=c(ax+by+cz)=cf(x)

    For (iii) I need to prove using all axioms that x(2,0,3) + y(-1,4,-2) is vector space.

    M1: a(u+v)=au+av
    a(u+v)=a((2x,0,3x)+(-y,4y,-2y))=a(2x-y,4y,3x-2y)

    au+av=a(2x,0,3x)+a(-y,4y,-2y)=a(2x-y,4y,3x-2y)

    M2: (a+b)u=au+bu

    But what is u and what is v here?
     
  14. Jun 17, 2009 #13

    Mark44

    Staff: Mentor

    Now I'm not sure. According to the function definition, f maps a vector (x, y, z) to a scalar ax + by + cz. IOW, f results in the dot product of a vector (x, y, z) with a fixed vector (a, b, c). If what I'm calling the fixed vector truly is fixed, then yes, there is just one function. If as you say, the vector contains constants but isn't fixed, then another function would be as you wrote it.
     
  15. Jun 17, 2009 #14
    CompuChip, now I see your post. Thanks for the help. Ok, I understand that I should go through all axioms, but what about (ii) and (iii) ??
     
  16. Jun 17, 2009 #15

    CompuChip

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    I assume you mean exercise (ii) and (iii)?
    Well, they work in the same way. For example, a vector is collinear with (1, -2, 3) if there exists a number r such that it can be written as r(1, -2, 3) = (r, -2r, 3r). So take two of them, for example r(1, -2, 3) and s(1, -2, 3) and show - among other things - that their sum can be written as t(1, -2, 3) for some number t.

    On one hand it is probably easier than you think, on the other hand it requires some level of abstraction. Where you probably used to think of a vector as an element of R3, that is: an arrow in space, you now have to think of a vector as "an element of a vector space." In particular instances, it may be an "arrow" in a space, or a function between to sets, or apples and oranges on which we define scalar multiplication and some binary operation that we call "addition".

    By the way, when you get more experienced you will be able to do these proofs more quickly. For example, you easily spot that a set is not a vector space because it is not closed under addition or does not contain a null vector. Or you quickly see that the set of mappings { f(x, y, z) = ax + by + cz | a, b, c in R } is "basically the same" (in mathematical terms: isomorphic) to R3 which is a vector space.
     
  17. Jun 17, 2009 #16
    CompuChip thanks for the reply.

    For the (iii) task probably x,y are scalars. So the vector has the form (2x,0,3x)+(-y,4y,-2y)=(2x-y,4y,3x-2y)
    So one vector u=(2x-y,4y,3x-2y). Should I take one more vector v=(2c-d,4c,3c-2d) and check the axioms?
     
  18. Jun 19, 2009 #17
    Am I right for (iii) task?

    I got (iv) task which says:

    "Is the set of all vectors (x,y,z) [itex]\in \mathbb{R}^3[/itex]:

    a) 2x+3y-z=0

    b)2x+3y-z+3=0

    for a) u=(x1,y1,z1) and v=(x2,y2,z2)

    u+v=(x1+x2,y1+y2,z1+z2)

    2(x1+x2)+3(y1+y2)-z1-z2=0

    2x1+3y1-z1 + 2x2+3y2-z2=0+0=0

    So its closed under addition.

    cu=c(x1,y1,z1)=(cx1,cy1,cz1)

    So its closed under scalar multiplication.

    But what if I need to prove all axioms?

    2x+3y-z=0

    [tex]x=\frac{z-3y}{2}[/tex]

    So the vector would be:

    [tex]u=(\frac{z-3y}{2},y,z)[/tex]

    Should I take one more vector v and prove all axioms?

    Thank you.
     
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