Question

Hello,
Can a subspace contain the zero element?

in example, the subspace w(a,b,0,d) where w belongs in R (a,b,c,d are constant numbers).
What would be the dimension of the above subspace?

Answer 1

A subspace must contain the zero element by definition. As for the example above, assuming (a, b, 0, d) is a basis for the subspace, the dimension must be less than or equal to 3 since you have at most 3 independent elements.

If (a, b, 0, d) isn't a basis, I have no idea about the dimension.

Answer 2

The zero element is in a subset (by definition)
Your elements are constant numbers (scalars?), so of zero dimension.

Answer 3

Agree with George, but still a bit confused by the question. Some more context would be helpful. You say that a, b, c, and d are constant numbers, which implies scalars, so I'm not sure how you're trying to build a subspace out of scalars? If they are vectors and we're in at least \(\mathbb{R}^3\), then it's what George said.

Answer 4

you can always test to make sure it is a subspace by making sure it has a zero element (which \(\mathbb R\) has already) and testing for closure under addition and scalar multiplication
as for the dimension I'm pretty sure it's 3

Answer 5

take, for instance \[]W=\{\langle x,y,c\rangle\}\]where \(c\) is a constant
that's clearly a 2 dimensional plane
I think you can extend that same argument here, but can't claim to be positive.

Answer 6

the above \(W\) as a subspace of \(\mathbb R^3\)

Answer 7

He says his elements are constant numbers in R

Answer 8

Yeah, the constant numbers bit is what's throwing me completely off.

Answer 9

In R3, you have 8 dims in total, 1,3,3,1 being 0,1,2,and 3 dim subspaces.

Answer 10

any constant though... so the terminology is confusing.
but clearly a,b, and c are distinguished from the 0 component by the fact that they can take on numerical values, so they act like variables in that sense...

Answer 11

the 0 component seems like the only real "constant"
we have to assume the other letters can take on any real value... I think

Answer 12

A subspace of R can only be a point or a line, right?

Answer 13

Right, and if it's a point it's the origin.

Answer 14

Well to put this clear,

We have a set of equations, and we get this matrix:
{ {9,5,7,4}
{9,5,3,4}
{8,1,2,4}
}

After Gauss elimination process we get this:
{ {1, 0, 0, 16/31}
{0, 1, 0, -4/31}
{0, 0, 1, 0}
}

So the solution set is: all the x,y,z,w that: w(-16/31, 4/31, 0 , 1) where w belongs in R.

I know the zero vector has to belong into the subspace..What I meant with the zero element is that the 3rd element here (-16/31, 4/31, 0 , 1) is "0"...

Doesn't that mean that the dimension of the solution set is R^3?

Answer 15

Yes but you need your set of scalars for scalar addition.

Answer 16

I stick by my answer above that this is the equivalent of \(W(x,y,0)\) which is a plane (dimesnion 2) so this should be 3 dimensional

Answer 17

So the numbers "-16/31, 4/31, 0 , 1" are called scalars?
Could you explain me what does the scalar addition have to do with the dim(solution_set)?

@nbouscal, so the dim(solution_set) = rank(of the matrix)

Answer 18

OK, I am in vectors and you are in matrices so we are at cross purposes somewhat...

Answer 19

Sorry, deleted my comment because I was wrong. Haha. Trying to remember linear algebra, haven't focused on it in a bit. The dimension of the solution space is equal to the dimension of the nullspace. In this case, that means a one-dimensional solution space? Sorry, trying to remember.

Answer 20

these scalar values put together in a 1x4 matrix form a vector (a non-scalar)
scalars are values that you can multiply the vectors by like \(c\langle x,y,0,z\rangle=\langle cx,cy,0,cz\rangle\)

Answer 21

in the above, c is a scalar

Answer 22

in any case, taking the 3rd component to be zero is like taking a cross-section of the otherwise 4 dimensional vector space

notice that if the third component were any constant other than zero, W would not be a subspace of \(\mathbb R^4\)

Answer 23

Okay. Hammered it out in my head for a minute, and I'm pretty sure this is what's going on. Your numbers -16/31, 4/31, 0, 1 form a solution to the equation Ax=0, which means that as a vector they form a basis for the nullspace of the matrix. The solution space is equal in dimension to the nullspace. So in this case, the solution space has dimension one.

Answer 24

I still don't have it straight in my head (I don't usually work with matrices).
Are we saying the solution is a line (intersection of 3 planes)?

Answer 25

Yeah, the solution space is a line. That line is formed by all scalar multiples of the vector [-16/31, 4/31, 0, 1] plus a particular solution that depends on what b is in Ax=b. Unless I'm doing it wrong, which is totally possible haha. Been a bit since I've done linear algebra.

Answer 26

OK, that sounds sensible, thanks:-)

Answer 27

So the w here, is a scalar, right?

And (w) it's a number in R?

@TuringTest: You said that "notice that if the third component were any constant other than zero, W would not be a subspace of R^4" , but, what subspace it would be then?

@nbouscal: So the rank of the matrix doesn't have to do anything with what I am trying to find, right?
And the only way to answer my question is to find the dimension of the kernel, which is one!

Answer 28

Actually, the rank of the matrix is related. The rank of a matrix plus the nullity of a matrix is equal to the number of columns of the matrix. So, since we have 4 columns and the rank is 3, we know that the nullity is 1, which in turn means that the solution space has dimension one because the solution space has the same dimension as the null space.

Answer 29

Thank you @nbouscal :)