Subsets & Subspaces: Determine 0 Vector & R4

[negation]

Homework Statement

a)Which of these subsets contain the zero vector 0 = (0, 0, 0, 0) ?
b)Which of these subsets are subspaces of R4 ?S = { (x1, x2, x3, x4) | x4 = -6 - 5 x1 }
T = { (x1, x2, x3, x4) | x4 is an integer }
U = { (x1, x2, x3, x4) | x1 + x4 ≤ -6 }

The Attempt at a Solution

If a set is closed under addition and closed under scalar multiplication, then it contains the zero vector. But isn't the definition of subset such that a vector space is a linear combination of all other vector. I'm terribly lost!

I'm really struggling badly with this. Can someone give me a leg up?

S = (x1,x2,x3,x4) and let W = (w1,w2,w3,w4)

then S+W= (x1+w1,x2+w2,x3+w3,x4+w4)

(x4+w4) = -6-5(x1+w1) (closed under addition?) (Where do I go from here?)

Let k = scalar

KW = (Kw1, kw2,kw3,kw4)

kw4 = -6-5(kw1) (closed under scalar?)

There are values for me to determine if the left hand side values equals the right hand side values.

 

[BvU]

Well,
a) fill in 0000:
S yes/no ?
T yes/no ?
U yes/no ?
b) Only have to look at the yesses ! Take a vector ##\vec x## and see if ##c\vec x## is always in the candidate.

Why are you looking at S + T ?
The question is: Is S a subspace of R4 or is it not
Or am I the one misinterpreting the exercise ?

 

[BvU]

By editing my reply becomes somewhat floating in the air!

Why don't you just check that 0000 is not in S and thereby discard S as a subspace?

 

[negation]

Well,
a) fill in 0000:
S yes/no ?
T yes/no ?
U yes/no ?
b) Only have to look at the yesses ! Take a vector ##\vec x## and see if ##c\vec x## is always in the candidate.

Why are you looking at S + T ?
The question is: Is S a subspace of R4 or is it not
Or am I the one misinterpreting the exercise ?

I'm equally confused too but I'm going with you.
If I get you correctly:

Let S = (x1,x2,x3,x4) = (0,0,0,0,)

S = (x1,x2,x3,x4)| x4=-6-5x1

x1 = 0, x2 = 0, x3=0,x4 = 0

x4 = -6-5(x1) = -6-5(0)=-6
-6=/= 0

0→ is not an element of S so S does not contain the 0→

Let T = (x1,x2,x3,x4) = (0,0,0,0)

T = (x1,x2,x3,x4)| x4 is an integer
x1=0, x2=0, x3=0, x4=0

I presume that by 'integer', it implies relative integer. 0 is a relative integer.
Hence (x4=0) = 0 = integer

The 0→ is an member vector in T and so T contains the 0→Let U = (x1,x2,x3,x4) = (0,0,0,0)

U = (x1,x2,x3,x4) | x1+x4 =< -6

x1 = 0, x2=0,x3=0,x4=0
x1 + x4 = 0, and so, 0 ~ -6 and ~< -6
0→ is not a member vector of U.

b) only T contains the zero vector.

Span(T) = R4 (?)

c1x1 + c2x2 + c3x3 +c4x4 = (x1,x2,x3,x4)
c4x4 = 0 since in part (a) it has been established x4 =0 and any scalar, c, times zero = 0.
Did I got it correctly?

 

[Mark44]

I'm equally confused too but I'm going with you.
If I get you correctly:

Let S = (x1,x2,x3,x4) = (0,0,0,0,)

No, S ≠ <0, 0, 0, 0>. S is a set of vectors in R⁴, subject to the condition you posted at the start of this thread. All you need to do is determine whether the zero vector satisfies the condition that x₄ = -6 - 5x₁

S = (x1,x2,x3,x4)| x4=-6-5x1

x1 = 0, x2 = 0, x3=0,x4 = 0

x4 = -6-5(x1) = -6-5(0)=-6
-6=/= 0

0→ is not an element of S so S does not contain the 0→

Rather than use "→" to stand for "vector" it's better to just write vector.

Let T = (x1,x2,x3,x4) = (0,0,0,0)

T = (x1,x2,x3,x4)| x4 is an integer
x1=0, x2=0, x3=0, x4=0

I presume that by 'integer', it implies relative integer. 0 is a relative integer.
Hence (x4=0) = 0 = integer

The 0→ is an member vector in T and so T contains the 0→


Let U = (x1,x2,x3,x4) = (0,0,0,0)

U = (x1,x2,x3,x4) | x1+x4 =< -6

x1 = 0, x2=0,x3=0,x4=0
x1 + x4 = 0, and so, 0 ~ -6 and ~< -6
0→ is not a member vector of U.

b) only T contains the zero vector.

Span(T) = R4 (?)

?
Span has nothing to do with showing that a subset of a vector space is a subspace of that vector space. Your textbook should show you the three things you need to check to demonstrate that a subset of a vector space is a subspace. Showing that the subset contains the zero vector is one of those things. What are the other two?

c1x1 + c2x2 + c3x3 +c4x4 = (x1,x2,x3,x4)
c4x4 = 0 since in part (a) it has been established x4 =0 and any scalar, c, times zero = 0.
Did I got it correctly?

 

[negation]

No, S ≠ <0, 0, 0, 0>. S is a set of vectors in R⁴, subject to the condition you posted at the start of this thread. All you need to do is determine whether the zero vector satisfies the condition that x₄ = -6 - 5x₁
Rather than use "→" to stand for "vector" it's better to just write vector.
?
Span has nothing to do with showing that a subset of a vector space is a subspace of that vector space. Your textbook should show you the three things you need to check to demonstrate that a subset of a vector space is a subspace. Showing that the subset contains the zero vector is one of those things. What are the other two?

One moment. Is part (a) correct?

1) zero vector must be a member vector of the set.
2) the set must be closed under addition operation
3) the set must be closed under scalar multiplication

How do I apply the above to the existing problem?

 

[BvU]

Yes part (a) is correct: Only T contains (0,0,0,0)
Now for part (b). We don't have to look at S and U any more, because they don't contain (0,0,0,0).
T only has as a criterion that x4 is an integer.
It satisfies 1): (0,0,0,0) is a member.
Now how about (2) ? If (a,d,c,d) and (e,f,g,h) both are members of T, what about the sum ?

 

[negation]

Yes part (a) is correct: Only T contains (0,0,0,0)
Now for part (b). We don't have to look at S and U any more, because they don't contain (0,0,0,0).
T only has as a criterion that x4 is an integer.
It satisfies 1): (0,0,0,0) is a member.
Now how about (2) ? If (a,d,c,d) and (e,f,g,h) both are members of T, what about the sum ?

Is coming up with (a,b,c,d) and (e,f,g,h) a general strategy to this sort of problems?

Let u = (u1,u2,u3,u4) and v = (v1,v2,v3,v4)

By property (2): u+v = (u1+v1,u2+v2,u3+v3,u4+v4)

x1 = u1+v1, x2 = u2+v2, x3 = u3+v3, x4 = u4+v4
then, since x4 is an integer, u4+v4 is an integer because u4+v4 = 0?

 

[BvU]

Almost. A general strategy is a bit presumptuous: it's more like a beginning of the solution process to take an example and then generalize while going along.

In this case if d and h are integer, then d+h is an integer too. So I am happy rule 2 holds.
You are happy too, not because u4+v4=0 but because if u4 is an integer and v4 is an integer too, then u4+v4 is also an integer. Which means that the sum vector is indeed IN the subset.

Now what about scalar multiplication ? Is criterion 3 satisfied ?

 

[negation]

Almost. A general strategy is a bit presumptuous: it's more like a beginning of the solution process to take an example and then generalize while going along.

In this case if d and h are integer, then d+h is an integer too. So I am happy rule 2 holds.
You are happy too, not because u4+v4=0 but because if u4 is an integer and v4 is an integer too, then u4+v4 is also an integer. Which means that the sum vector is indeed IN the subset.

Now what about scalar multiplication ? Is criterion 3 satisfied ?

Suppose U = (u1,u2,u3,u4) and k is a scalar an element of real number.

kU = (ku1,ku2,ku3,ku4)

x4 = ku4
Since k is a scalar an element of real number and u4 an integer, then ku4 is an integer.

 

[LCKurtz]

Suppose U = (u1,u2,u3,u4) and k is a scalar an element of real number.

kU = (ku1,ku2,ku3,ku4)

x4 = ku4
Since k is a scalar an element of real number and u4 an integer, then ku4 is an integer.

Are you actually thinking about this?

 

[BvU]

Back to the basic strategy: fill in something and see where it leads:

(0,0,0,5) is a member. 1.5 is a scalar. 1.5 (0,0,0,5) = (0,0,0,7.5) So here we have found a counter example. One case where a scalar times a member does not give us a member. Conclusion ?

 

[negation]

Back to the basic strategy: fill in something and see where it leads:

(0,0,0,5) is a member. 1.5 is a scalar. 1.5 (0,0,0,5) = (0,0,0,7.5) So here we have found a counter example. One case where a scalar times a member does not give us a member. Conclusion ?

Where did 0,0,0,5 came from?

 

[BvU]

I invented it. It is a member of T. Anything else would have been fine too. Point is that if the fourth coordinate is an integer and you multiply by a scalar that is not an integer, then you get something with a fourth coordinate that is not an integer and that means you are no longer in the subset. That means the subset is not a subspacce.

 

[negation]

I invented it. It is a member of T. Anything else would have been fine too. Point is that if the fourth coordinate is an integer and you multiply by a scalar that is not an integer, then you get something with a fourth coordinate that is not an integer and that means you are no longer in the subset. That means the subset is not a subspacce.

Yes you're right!
I forgotten that it was real number, k x integer.
It's terrible having only 4 hours of sleep each day for the past 2 weeks..

 

[BvU]

Get some rest. We won't go away all at the same time, so someone's there when you get up and have more questions !