About the definition of the dimension DEFINITION: The dimension of a space is the number of vectors in every basis.
This is the definition of dim from Introduction to Linear Algebra 4th by Strang.
I have a question here. A vector in $R^n$ is belong to n-dim space.  When we talk about a plane, we refer to the 2-dim subspace in 3-dim space or just the 2-dim space? If we agree with the former, doesn't it belong 3-dim space? So why is the statement The dimension of a space is the number of vectors in every basis valid?
I think it a bit difficult because of the language, haha, thanks if you help!
 A: That definition isn't worded correctly. It should be "The dimension of a vector space is the number of vectors in every basis of that space, containing only vectors from that space."
Therefore, when we talk about a plane, which is a 2-dimensional subspace in a 3-dimensional vector space, the bases of vectors only from that plane only have 2 vectors in them, so it can still be 2-dimensional without contradicting that the original vector space had 3 vectors in their bases.
Let's take an example. $\Bbb{R}^3$  will be our 3-dimensional vector space and $\{(x, y, z) \in \Bbb{R}^3 \mid z=0\}$ will be our 2-dimensional vector space. (Our 2-dimensional subspace is the same as the $xy$-plane.)
It is very easy to show that the following is a basis for $\Bbb{R}^3$:
$$\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$$
However, in our two-dimensional subspace, $z$ is always $0$. Therefore, the vector $(0, 0, 1)$ is not part of our basis because it has a non-zero $z$-coordinate. $(0, 0, 1)$ is not in the plane, so it can not be in the plane's basis. The other two vectors are, however, because there is no restriction on the $x$ and $y$-coordinates. Therefore, the plane has the following basis:
$$\{(1, 0, 0), (0, 1, 0)\}$$
Thus, the whole vector space has a basis with three elements, but the plane is only a subset of the vector space, so it only contains a smaller basis. This is how a 2-dimensional subspace can be inside of a 3-dimensional subspace.
