Question on finding the dimension and basis of a family of vectors

Question

Let's say I have the following family

$f = ((1,0,1,0),(-1,0,-1,0), (1,1,1,1),(0,1,0,1), (1,2,1,0))$.

I want to first find its rank. I can say, by eliminating all the colinear vectors, that the rank of my family is equal to

rank of $f = \dim(span((1,0,1,0),(0,1,0,1), (1,2,1,0))) = 3$.

Now, I'm very confused about the the whole idea of dimensions. When someone say that the dimension of a basis is always equal to the dimension of the vector space, they're speaking about the number of vectors in the basis, correct? Does this mean that a family of $3$ vectors can be a basis, even if the vectors in the family are of dimension $4$.

In the particular example I gave, can I say that $((1,0,1,0),(0,1,0,1),(1,2,1,0))$ is a basis of $\mathbb R$$^3$? Or does that make no sense? Thank you.

Answer 1

The dimension of a finite-dimensional vector space is the number of vectors in a basis for the vector space. This number is independent of the choice of the basis.

You can't say that "$\{(1,0,1,0),(0,1,0,1),(1,2,1,0)\}$ is a basis of $\Bbb{R}^3$" because basis vectors are "representatives" of the vector space, so they must live in that vector space. However, in this numerical example, the proposed vectors live in $\Bbb{R}^4$. You may rather say that this set of linear independent vectors spans a three-dimensional subspace of $\Bbb{R}^4$.

Answer 2

$\mathbb R$$^3$ has dimension 3, but that doesn't mean that a space with dimension 3 is $\mathbb R$$^3$. All vector spaces of finite dimension are isomorphic to all vector spaces of the same dimension (given a fixed scalar field), but they aren't equal. While each finite dimension has a canonical vector space, and we can speak of "the" vector space of that dimension, we can do so only up to isomorphism. So ((1,0,1,0),(0,1,0,1),(1,2,1,0)) is the basis of a space that is isomorphic to $\mathbb R$$^3$, but since the vectors aren't in $\mathbb R$$^3$, the space is not equal to $\mathbb R$$^3$. There are multiple 3 dimensional subspaces of $\mathbb R$$^4$, and they are all isomorphic to each other, but they are not equal to each other.