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Vector calc, answer check

  1. Dec 21, 2008 #1
    Hi all, i'm currently revising for a second year moduel by looking through past exam
    papers, i dont have the answers to these papers but i do have similar questions so i'm
    trying to piece together the answers from what i do have, i think this is right but would
    be greatful if someone could check it through so i'm not revising the wrong thing for the
    next few weeks, thanks

    1(a) Let V be a vector space over field K. What is meant by saying that W is a vector
    subspace of V?

    my answer: Assuming W as a subset of V, W is a subspace if and only if the following 3
    criteria are satisfied

    (I)W contains the zero vector
    (II)if uand vare elements of W then so is (u+v)
    (III) If u is an element of W and c is a scalar from K then cu is also an
    element of W

    (b) Which of the following are vector supspaces of R3? Give proofs or
    counterexamples. as appropriate.

    (i) W1 ={(x1,x2,x3E R3) :
    x1 + 5x2 + 6x3=0

    (I)If (x1,x2,x3) E W1 and (x4,x5,x6) E W1

    so that x1 + 5x2 + 6x3=0
    and x4 + 5x5 + 6x6=0
    then (x1+x4,x2+x5,x3+x6)
    lies in this space since we get
    (x1+x4) +5(x2+x5) +6(x3+x6)= (x1+5x2+6x3)+(x4+5x5+6x6)=0

    (III) If (x1,x2,x3) E W1 and c E R ( is this R or R3)?
    then x1+5x2+6x3=0 implies that cx1+5cx2+6cx3=0c

    therefore c(x1+5x2+6x3)=0
    therefore C0=0 therefore W1 is a subspace of R3 as this holds also

    (ii)W2 ={(x1,x2,x3E R3) :
    x1,x2,x3=(0,0,0) satisfies x1*x2*x3=0
    (II)if (x1,x2,x3 E W2) and
    (x4,x5,x6 E W2
    so that x1*x2*x3=0 and x4*x5*x6=0
    then (x1+x4)*(x2+x5)*(x3+x6)=(x1*x2*x3)+(x4*x5*x6)=0
    this is not the case as (1,0,0) and (0,2,2) would both satisfy x1*x2*x3=0 but together would give (x1+x4)*(x2+x5)*(x3+x6)+4 therefore W2is not a subspace of r3

    thanks in advance for checking it through hopefully its right

    Few more questions i've attempted now, pritty sure these are right but a 2 minute check over cant hurt right? thanks again
    consider the following subspaces of R3

    U={(a+3b,a,b) : a,b E R V={(c,0,0) : c E R W={(4d,d,d) : d E R

    giving brief reasons, determine whether or not
    (i)R3 =Udirect sumV
    (ii)R3=Udirect sumW
    (iii)R3=Vdirect sumW

    only done the first one so far but will update/edit as i complete the others

    (i) check that VnW=(0,0,0), if (x,y,z)=(a+3b,a,b)=(c,0,0)
    then we have x=a+3b,y=a=0,z=b=0
    this gives a=0,b=0 and therefore c=0
    i.e. (x,y,z)+(0,0,0)
    next we show every vector (x,y,z) can be writen w+v with vE V and w E W
    we need to solve x=a+3b+c
    and we can getting

    i beleive that the mere fact that these solve shows that they are a equal to R^3? correct me if i'm wrong thanks :)

    (ii) bit stuck on this one or might just not be sure what my answer shows
    first check that UnW=(0,0,0) i dont think it does because i get
    then x=a+3b=4d,y=a=d,z=b=d
    which i cant make a zero out of so i cant get (x,y,z)=(0,0,0) so i beleive this shows me not only that R^3doesnt equal Udirect sum W but it also shows that there not a direct sum in the first place?
    just took another look at it now and i'm almost convinced i'm right with (4,1,1) being an example of a non zero vector?

    without writing it all out at the end i get d=z=y and c=x-4d so i get c=x-4y or c=x-4y, does this show that it does work as the equations are solvable or does the d=z=y alter things some how?
    Last edited: Dec 21, 2008
  2. jcsd
  3. Dec 21, 2008 #2


    User Avatar
    Homework Helper

    Looks ok. You could also combine II and II by noting that [tex]au+bv \in W[/tex] if given a,b are any elements of K and u,v are any vectors in V.
    It's R.
    You should also show that the zero vector is also present in this subspace, for which x1=x2=x3 = 0.

    Yes, that is a correct counter-example.

    You meant UnV, right.

    Yes, ok to me.

    Yes, looks ok. You've shown how to find values of a,b,c (and hence vectors belonging to U and V) such that any (x,y,z) in R3 is expressible in terms of these. And as above, you meant U,V not W I guess.

    If you wrote the three subspaces U,V,W as the linear span of some vectors you should be able to see that the sole vector in the basis of W is a linear combination of the other two vectors in the basis of U. Then clearly it can't be a direct sum since [tex]U \cap W \neq {\mathbf{0}} [/tex].

    Again it helps you have already expressed the three subspaces in terms of linear spans of some specified basis. Then take note of the number of linearly indepedent vectors in [tex]V\oplus W[/tex].
  4. Dec 22, 2008 #3
    thanks mate, most helpful
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