Thread: Dimension of Range and Relations

If we define the linear map T: Rn --> Rm and x --> Ax and the dimension of the nullspace of A is 2, what is the dimension of the range of A? What can you say about the relation between n and m?

I think the dimension of the range of A is m-2 because dim(nullspace(A)) + dim(range(A)) = dim(A) and that would mean m>n but I could be understanding this wrong.

I don't think you're quite applying the rank-nullity theorem correctly here. That theorem says that if you have a linear transformation then

not

How does this change your result?

Oh I didn't realize that the rank-nullity theorem would apply to Rn. And kernel is equivalent to nullspace, right?, so dim(range(T)) = n-2. So does this mean that m<n because if the dimension of the range is less than n, then that means that the dimension of the codomain is less than the dimension of the domain?

Thank you for your help.

Kernel = nullspace. That is correct.

Your figure of n - 2 for the range is correct, as well.

The dimension of the codomain is, I believe, always less than or equal to the dimension of the domain. You can think of this as a "conservation of information" theorem, almost. A linear transformation can't produce new information, it can only manipulate existing information, or, if the rank of the matrix is less than the number of rows, it can destroy information.

Just remember that if you have a linear transformation then

and



Make sense?

Oh ok. So what I'm getting from this is that m<=n but also that n-2<=m since rank is less than the dimension of the domain and codomain so m is in between n-2 and n?

Well, I think you have to distinguish between the range and the co-domain. See here for an explanation. The fact that vectors go from Rn to Rm via T does not mean that T is onto Rm (that is, every vector in Rm can be "hit" by a vector coming from Rn via T). How are you defining range and codomain? And what does the notation



mean to you? Does it mean automatically that is onto