Base, number of dimensions, subspace, dimensionality drop

First, the question of the proposed

It is difficult to understand the difference between the length of the base (a set of linearly independent vectors) and the dimensionality by the influence of different dimensions of space, plane and line. The length of the base = number of dimensions?

To know the representation of the space, the base is three degrees of freedom, and the plane is two degrees of freedom. In the projection is the dimensionality of the drop ...

It looks very chaotic!!

Second, the analysis of the problem

First analyze several conclusions:

(1) Dimensions of the subspace ≤ the dimensions of the original space

Since the set of subspace is a subset of the original set of spaces, there is no doubt that the number of linearly independent vectors required by the subspace is ≤ the number of linearly independent vectors required by the original space, so the conclusion is proved.

(2) The length ≠ dimension of the base

To give a counter example, obviously (a1,a2,0), satisfies the addition and multiplication operation sealing, is the subspace of three-dimensional space, it needs two linearly independent vector representation, therefore is the dimension number is two. But its base length is 3.

Reference documents

Base, number of dimensions, subspace, dimensionality drop