1. Dec 21, 2008

### franky2727

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
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
y=a
z=b
and we can getting
b=z
a=y
c=x-y-3z

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
(x,y,z)=(a=3b,a,b)=(4d,d,d)
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?

(iii)
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. Dec 21, 2008

### Defennder

Looks ok. You could also combine II and II by noting that $$au+bv \in W$$ 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 $$U \cap W \neq {\mathbf{0}}$$.

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 $$V\oplus W$$.

3. Dec 22, 2008