Yes, a subspace is a space in itself. The conditions required for a subspace $+$ the conditions that the subspace inherits from the space make it a space in it's own right. You should check here. Why are they important? A subspace lives inside a space. The fact that it has in itself all the nice properties of a space make it interesting.

Mar 31, 2014 · Actually, there is a reason why a subspace is called a subspace: It is also a vector space and it happens to be (as a set) a subset of a given space and the addition of vectors and multiplicataion by scalars are "the same", or "inherited" from that other space.

The span of a single non-zero vector is a one-dimensional subspace. The span of two linearly independent vectors is a two-dimensional subspace. And so on. It is a convenient way to talk about "the set of all linear combinations you can get by starting out with a certain set of vectors." Or, in fewer words: the subspace generated by a set of ...

Oct 12, 2011 · The union of two subspaces is a subspace if and only if one of the subspaces is contained in the other. The "if" part should be clear: if one of the subspaces is contained in the other, then their union is just the one doing the containing, so it's a subspace. Now suppose neither subspace is contained in the other subspace.

Mar 12, 2019 · is a subspace. I know I have to check for the zero vector, addition and scalar multiplication. Here lies my question however. If I use $(0,0,0)$ to check if the zero vector exists, it clearly does. However, if I multiply $(0,0,1)$ by some scalar, say, $50$, scalar multiplication does not hold so this isn't a subspace. Here lies my problem.

Jul 21, 2014 · Subspace is a definition coming from vector space and I take you know what dimension of vector space means. Anything contained in $\mathbb{R}^n$ is a subset of it so notion of dimension don't really exist $\endgroup$

Dec 2, 2017 · A subspace of a space with a countable base also has a countable base (the intersections of the countable base elements with the subspace), and a subspace with a countable base is separable (pick an element from each non-empty base element).

Firstly, there is no difference between the definition of a subspace of matrices or of one-dimensional vectors (i.e. scalars). Actually, a scalar can be considered as a matrix of dimension $1 \times 1$. So as stated in your question, in order to show that set of points is a subspace of a bigger space M, one has to verify that :

Sep 9, 2015 · The geometrical meaning of a subspace of a three dimensional space being a two dimensional space is that all the vectors from that subspace are contained on a plane in the three dimensional space - besides the meaning of needing only 2 coordinates do be uniquely defined even on a three dimensional space, because the third coordinate is defined ...

Mar 25, 2018 · $\begingroup$ I haven't learned what a subspace topology was, I think I meant to just say subspace in the title. The problem I was having was resolved once I realised that, for example, $[-1,1]$ is the largest open subset of the space $[-1,1]$.