Understanding Vector Spaces: Definition and Examples | Vector Calc Answer Check

[franky2727]

Hi all, I'm currently revising for a second year moduel by looking through past exam
papers, i don't 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 can't 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 written 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 believe 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 don't 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 can't make a zero out of so i can't get (x,y,z)=(0,0,0) so i believe 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?

 

[Defennder]

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

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.

(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)?

It's R.

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

You should also show that the zero vector is also present in this subspace, for which x₁=x₂=x₃ = 0.

(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

Yes, that is a correct counter-example.

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

(i) check that VnW=(0,0,0)

You meant UnV, right.

, 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)

Yes, ok to me.

next we show every vector (x,y,z) can be written 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 believe that the mere fact that these solve shows that they are a equal to R^3? correct me if I'm wrong thanks :)

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.

(ii) bit stuck on this one or might just not be sure what my answer shows
first check that UnW=(0,0,0) i don't 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 can't make a zero out of so i can't get (x,y,z)=(0,0,0) so i believe 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?

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}}.

(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?

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.

 

[franky2727]

thanks mate, most helpful