Proving vectors are in the column space

[Taylorw369]

How would you prove that adding two vectors in the column space would result in another vector in the column space?

I know this is maybe the most basic property of vectors and subspaces, and that the very definition of the column space says it's spanned by vectors in the column space. Is there any way to prove this, though?

 

[HallsofIvy]

How would you prove that adding two vectors in the column space would result in another vector in the column space?

I know this is maybe the most basic property of vectors and subspaces, and that the very definition of the column space says it's spanned by vectors in the column space. Is there any way to prove this, though?

What, exactly, are you asking? You say that you know "very definition of the column space says it's spanned by vectors in the column space." You understand that you don't "prove" definitions don't you?

 

[WWGD]

The column space is a subspace, so that it is closed under addition of vectors.

EDIT: Sorry, I guess you were trying to show that the column space is a subspace. Let me
think it through.

EDIT: by definition, see , e.g.:http://en.wikipedia.org/wiki/Column_space , the column space

is the set of all linear combinations of the row vectors. Take then a combination of row

vectors and add another combination to it to show the sum is itself a combination.

Does that help?

EDIT 2: By definition, a linear transformation takes vector spaces (subspaces) to subspaces.
The column space is the linear image of a vector space is a vector space; linear maps take
vector spaces to vector spaces.

 

[Fredrik]

If X is a vector space over ℝ and S is a subset of X, the set of linear combinations of elements of S is a subspace of X. The proof is trivial: Let's denote the set of linear combinations of elements of S by W. We obviously have $0\in W$. Let $a,b\in\mathbb R$ and $x,y\in S$ be arbitrary. We have $ax+by\in W$. (That's the entire proof).

The subspace W can be equivalently defined as the intersection of all subspaces of X that contain S. If you show that these definitions are equivalent, you can easily prove that W is the "smallest" subspace of X that contains S, in the sense that if V is another such subspace, we have $W\subseteq V$. You can also easily prove that there's only one "smallest" subspace of X that contains S.

So W is the unique smallest subspace of X that contains S. This space is called the subspace generated by S, or the subspace spanned by S.