What Are the Requirements for a Subset to Be Considered a Subspace?

[Krovski]

I'm having trouble conceptualizing exactly what a subspace is and how to identify subspaces from vector spaces.

I know that the definition of a subspace is:
A subset W of a vector space V over a field $\textbf{F}$ is a subspace if W is also a vector space over $\textbf{F}$ w/ the operations of vector addition and scalar multiplication.

So if I have to define a subspace of $\textbf{R^3}$, is it enough to show that the new vector, say W, exists in $\textbf{R^3}$?

 

[tiny-tim]

Hi Krovski!

A subset W of a vector space V over a field $\textbf{F}$ is a subspace if W is also a vector space over $\textbf{F}$ w/ the operations of vector addition and scalar multiplication.

So if I have to define a subspace of $\textbf{R^3}$, is it enough to show that the new vector, say W, exists in $\textbf{R^3}$?

I don't understand your question …

W is a space (a line or a plane), not a "new vector". ​

 

[Krovski]

this is exactly why I need help understanding what a subspace is lol.

so would i then say I need to show that W is also a vector space?

and that it all its elements also belong in V?

or am I thinking of subsets?

 

[tiny-tim]

W must be a subset of V,

which means that every element of W is in V

also, W must be closed under vector addition and scalar multiplication

 

[Krovski]

ok thank you.

more precisely,

if W=(a1,a2,a3)
would W be a subspace of $\textbf{R^3}$
if, say, we had one element a1=a2+2?

then W=(a2+2,a2,a3)

it technically isn't seen as c(a1,a2,a3) or even adding another vector in $\textbf{R^3}$
so that this vector, we'll call, (b1,b2,b3) means
(a1+b1,a2+b2,a3+b3) =/= (a2+2,a2,a3)
and therefore W is not a subspace?

 

[tiny-tim]

if W=(a1,a2,a3)

you mean, if W is a point?

then W is not closed under scalar multiplication (unless W is (0,0,0))

(and I'm off to bed :zzz:)

 

[Krovski]

you mean, if W is a point?

then W is not closed under scalar multiplication (unless W is (0,0,0))

(and I'm off to bed :zzz:)

to clarify,
W=(a1,a2,a3)$\epsilon$ ℝ^3 (this set is a subspace in ℝ^3)


so would
W1=(a2+2,a2,a3) also be a subsace in ℝ^3? or does that violate the definition of a subspace with the a2+2 term?

 

[DonAntonio]

to clarify,
W=(a1,a2,a3)$\epsilon$ ℝ^3 (this set is a subspace in ℝ^3)


so would
W1=(a2+2,a2,a3) also be a subsace in ℝ^3? or does that violate the definition of a subspace with the a2+2 term?

You need to be careful how you write mathematical stuff. It seems to be you actually mean $$W:=\{(a_1,a_2,a_3)\in\mathbb{R}^3 \, / \,a_1=a_2+2\}$$
Assuming this, then no: W is not a subspace as it doesn't contain the zero vector, which is a necessary condition to be a subspace.

DonAntonio

 

[HallsofIvy]

I recommend you read an introductory text on linear algebra. It would go over the basic definitions in much more detail that we can here.

If you already know what a vector space is then a "subspace" is a subset of a vector space that also satisfies all the conditions for a vector space. However, many of those conditions, such as "u+ v= v+ u" are true for any vectors in the vector space so don't need to be checked again. What you do need to check is that
1) the set is not empty.
2) the set is closed under addition: The sum of two vectors in the set is again in the set.
3) the set is closed under scalar multiplication: the product of any scalar (member of the underlying field) with a vector in the set is again in the set.

For $\{(a_2+ 2, a_2, a_3)$ we can note that (2, 0, 0) is in the set because [/itex]a_1= 2= 0+ 2= a_2+ 2[/itex] but the product of the scalar 3 and (2, 0, 0) is (6, 0, 0) which is NOT in that set: $6\ne 0+ 2$.

The condition Don Antonio refers to, that the 0 vector must be in the subset in order that it be a subspace, follows from the conditions above:
(1), that subset be non-empty, requires that some vector, v, be in the subset. (3), that the subset be closed under scalar multiplication, requires that (-1)v be in the set. (2), that the subset be closed under addtion, requires that v+ (-1)v= (1+ (-1))v= 0v= 0, the zero vector, be in the set.

 

[Krovski]

I recommend you read an introductory text on linear algebra. It would go over the basic definitions in much more detail that we can here.

If you already know what a vector space is then a "subspace" is a subset of a vector space that also satisfies all the conditions for a vector space. However, many of those conditions, such as "u+ v= v+ u" are true for any vectors in the vector space so don't need to be checked again. What you do need to check is that
1) the set is not empty.
2) the set is closed under addition: The sum of two vectors in the set is again in the set.
3) the set is closed under scalar multiplication: the product of any scalar (member of the underlying field) with a vector in the set is again in the set.

For $\{(a_2+ 2, a_2, a_3)$ we can note that (2, 0, 0) is in the set because $\{(a_2+ 2, a_2, a_3)$ but the product of the scalar 3 and (2, 0, 0) is (6, 0, 0) which is NOT in that set: $6\ne 0+ 2$.

The condition Don Antonio refers to, that the 0 vector must be in the subset in order that it be a subspace, follows from the conditions above:
(1), that subset be non-empty, requires that some vector, v, be in the subset. (3), that the subset be closed under scalar multiplication, requires that (-1)v be in the set. (2), that the subset be closed under addtion, requires that v+ (-1)v= (1+ (-1))v= 0v= 0, the zero vector, be in the set.

Yes I have been trying to read this one text, but I'm definitely behind the power curve for this topic since I haven't had much experience or luck doing proofs. I have to think of everything graphically which can become hard to do when dealing in higher dimensions.

So, what I have seen from all of your posts, and thank you all very much, is that because of the (a$_{2}$+2) term, the subset of W is not closed under scalar multiplication. If (a$_{n}$)=0 the vector W = (2,0,0) which is in the subset of W. BUT, this is where I'm a little lost, since (2,0,0) is in the subset of W that also means that it too should follow the rules of the subspace (i.e. closed under scalar multiplication, vector addition and non-emptiness) which is why 3(2,0,0) = (6,0,0) but that since $\{(a_2+ 2, a_2, a_3)$ we get the new term 6 but that doesn't follow the specifics of this subpace considering $\{a_2+ 2 = a_1)$ : meaning 6 is not 0+2.

So W is not closed under scalar multiplication.