7. Infinite dimensional spaces, sums and intersections

Infinite system of vectors

We say that a vector v is a linear combination of an infinite system of vectors U, if there exists a finite subset of U, such that v is its linear combination. A system of vectors is linearly independent, if none of its vectors is a combination of rest of them. It is a basis of a space V, if it is linearly independent and every vector of V is its linear combination.

E.g., system \{e_i\colon i\in \mathbb{N}\}, where e_i is the infinite sequence of zeroes with one only on i-th place is linearly independent in the space of all real sequences \mathbb{R}^\infty. But it is not a basis of this space, since the sequence of infinitely many ones is not a linear combination of this system (notice the finite subset part of the definition of linear combination). But it is a basis of space \mathbb{R}^\infty_c of all sequences which have zeroes from some moment on.

Infinitely dimensional spaces

A space is infinitely dimensional, if it has no basis consisting of finitely many vectors. By Zorn Lemma (see here), every space has a basis, so an infinite dimensional space has a basis consisting of infinite number of vectors (sometimes even uncountable).

It is worth noticing, that also in an infinite dimensional space the coordinates of a vector with respect to a given basis are unique, and a basis is a maximal linearly independent system of vectors, as well as minimal system of vectors spanning the considered space.

Intersection of subspaces

If V and W are subspaces of a space X, then V\cap W is a subspace of X as well.

E.g. if V=\text{lin}((1,0,1,1),(0,1,0,0)), W=\text{lin}((2,0,2,2),(0,0,1,0)), then V\cap W=\text{lin}((1,0,1,1)). If on the other hand V=\{(x,y,z)\colon x+y-z=0\} and W=\{(x,y,z)\colon 2x+2y-z=0\}, then V\cap W is the space of solutions to

    \[\begin{cases} x+y-z=0,\\2x+2y-z=0.\end{cases}\]

Sum of subspaces

If V and W are subspace of X, then V+W=\{v+w\colon v\in V\land w \in W\} is also a subspace of X.

E.g. if V=\text{lin}((1,0,1,1),(0,1,0,0)), W=\text{lin}((2,0,2,2),(0,0,1,0)), then V+ W=\text{lin}((1,0,1,1),(0,1,0,0),(0,0,1,0)).

It is easy to notice that V+W=\text{lin}(V\cup W) and that in the case of finite dimensional spaces \dim(V+W)=\dim V+\dim W-\dim(V\cap W).

Direct sum

We say that X is a direct sum of its subspaces V and W (denoted by X=V\oplus W), if X=V+W, but also V\cap W=\{0\}. In other words, every vector \alpha \in X can be uniquely described as v+w, where v\in V and w\in W. Obviously, if X=V\oplus W, then \dim X=\dim V+\dim W.

E.g. \mathbb{R}^2=\lin((1,0))\oplus\lin((1,1)), but even though \mathbb{R}^2=\lin((1,0),(1,1))+\lin((1,1)), this sum is not direct.