Column space and nullspace relationship?

[kostoglotov]

I have just been studying Nullspaces...

I want to make the following summary, will it be correct?

C(A) is all possible linear combinations of the pivot columns of A.

N(A) is all possible linear combinations of the free columns of A (if any exist).

edit: I have a feeling these are insufficient as definitions, but I'll leave it as it is for now.

edit2: I just realized that the second part could give the wrong dimensions for N(A).

 

[HallsofIvy]

For a linear transformation, A, from vector space U, of dimension m, to vector space V, of dimension n, the "null space of A" is the subspace of U such that if v is in U, Au= 0. The "column space of A", the span of the columns when A is written as a matrix, is the subspace of V spanned by the columns written as vectors. There is no direct relation between those two spaces (as, one is a subspace of the other) because one is subspace of U and the other is a subspace of V. There is however the "dimension law", that the dimension of the column space (the "rank" of A) plus the dimension of the null space (the "nullity" of A) is equal to the dimension of U. That can be shown by writing a basis that contains a basis for the null space itself plus other vectors that then map to the column space.

 

[kostoglotov]

For a linear transformation, A, from vector space U, of dimension m, to vector space V, of dimension n, the "null space of A" is the subspace of U such that if v is in U, Au= 0. The "column space of A", the span of the columns when A is written as a matrix, is the subspace of V spanned by the columns written as vectors. There is no direct relation between those two spaces (as, one is a subspace of the other) because one is subspace of U and the other is a subspace of V. There is however the "dimension law", that the dimension of the column space (the "rank" of A) plus the dimension of the null space (the "nullity" of A) is equal to the dimension of U. That can be shown by writing a basis that contains a basis for the null space itself plus other vectors that then map to the column space.

Yeah, I just realized that combining the columns of A could easily give a vector of the wrong dimensions for N(A).

 

[kostoglotov]

For a linear transformation, A, from vector space U, of dimension m, to vector space V, of dimension n, the "null space of A" is the subspace of U such that if v is in U, Au= 0. The "column space of A", the span of the columns when A is written as a matrix, is the subspace of V spanned by the columns written as vectors. There is no direct relation between those two spaces (as, one is a subspace of the other) because one is subspace of U and the other is a subspace of V. There is however the "dimension law", that the dimension of the column space (the "rank" of A) plus the dimension of the null space (the "nullity" of A) is equal to the dimension of U. That can be shown by writing a basis that contains a basis for the null space itself plus other vectors that then map to the column space.

So the nullspace is a subspace of the column space...that makes sense.

 

[HallsofIvy]

So the nullspace is a subspace of the column space...that makes sense.

? That's exactly the opposite of what I said in what you quote! I said that if A maps vector space U to vector space V, the column space is a subspace of V and the null space is a subspace of U so that there is NO necessary relationship between the two. I said and you quote: "There is no direct relation between those two spaces (as, one is a subspace of the other) because one is subspace of U and the other is a subspace of V."

 

[WWGD]

This : https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra

may help clarify the relations.