I am used to work with vector spaces that have a basis of the same dimension. For example, in quantum mechanics one can always represent the Hamiltonian in some basis (of dimension $n$), diagonalize it (suppose that the spectrum is no degenerate), and then take the set of eigenvectors as another good basis (again of dimension $n$).
I have, however, the following Lemmas on which I want to formulate my question.
L1. Suppose that the vectors $x_1,\dots,x_n$ span the linear space $X$ and that te vectors $y_1,\dots,y_j$ in $X$ are linearly independent. Then $j\leq n$
L2. A linear space $X$ which is spanned by a finite set of vectors $x_1,\dots,x_n$ has a basis.
My first question is that if I can have a linear space $X$ of dimension $\operatorname{dim}X$, s.t. it can be spanned by $m$ vectors with $m\geq\operatorname{dim}X$. This can never happen in $\mathbb{R}^n$ right? In this case, we always have (from L1) that $n=j$.
No it also happens in $\mathbb{R}^n$. Consider all the vectors in $\mathbb{R}^n$, which is uncountably many, then they span $\mathbb{R}^n$. The thing is that a $n$-dimensional vector space $V$ can be spanned by $n$ (necessarily linearly independent) vectors, but it can also be spanned by more than $n$ (necessarily linearly dependent) vectors.
Edit : Simple example in $\mathbb{R}^2$ is the set $\{(1,0),(0,1),(1,1)\}$. When you have the first two, then you don't need the third one. But these three vectors still span $\mathbb{R}^2$ which is $2$-dimensional.