How many bases can a vector space have? Hello guys I am currently taking a linear Algebra class; I stumbled upon this question:
How many bases can a vector space have? Are they unique?
Thanks in advance.
 A: This is a very interesting question. 
It depends on the base field, but let's assume you have a (normed) real vector space. 
If it's $\mathbb R^2$ then there are many, many possible bases. For example $\langle (x,0),(0,y):x,y \in \mathbb R, xy \neq 0\rangle $ are all possible sets of basis vectors.  The full set of basis vectors is $\langle (a,b),(c,d):ad-bc \neq 0\rangle$.
A more interesting question is: what does the set of all bases "look like"?
If you consider the set of all ordered bases, i.e. you think of orientation so that $\langle (1,0),(0,1)\rangle$ is different to $\langle (0,1),(1,0) \rangle$ then the set of all bases for a two dimensional real vector space $\langle (a,b),(c,d):ad-bc \neq 0\rangle$ is just $\mathbb R^4$, but with the surface $ad-bc = 0$ cut out.
What if you consider the set of orthonormal bases of $\mathbb R^n$, i.e. basis vectors have unit length and are mutually perpendicular. You can start with the standard basis: 
$$\langle (1,0,\ldots,0),(0,1,0,\ldots,0),\ldots,(0,\ldots,0,1,0),(0,\ldots,0,1)\rangle$$ Any new orthonormal basis can be made a linear transformation for which angles and lengths are preserved. If angles and lengths are preserved then it will still have unit vectors that are mutually perpendicular. The set of all linear transformations that preserves angles and lengths is called the orthogonal group and denoted by $\mathrm O(n)$.
The orthogonal group $\mathrm O(n)$ is a compact Lie group of dimension $\frac{1}{2}n(n-1)$.
In the case of all orthonormal bases of $\mathbb R^2$ we get $\mathrm{O}(2)$ which has dimension $\frac{1}{2}\cdot 2\cdot(2-1) = 1$. 
In the case of all orthonormal bases of $\mathbb R^3$ we get $\mathrm{O}(3)$ which has dimension $\frac{1}{2}\cdot 3\cdot(3-1) = 3$.
The topology or "shape" of these spaces is very interesting. In the case of $\mathbb R^2$, the set of all sets of bases $\mathrm{O}(2)$ is two disjoint circles $S^1 \times \{-1,1\}$. To go from the standard basis $\langle (1,0),(0,1)\rangle$ to any other orthonormal basis might take a simple rotation or first a reflection and then a rotation.
A: Every nonzero number is a basis for the one dimensional real vector space $\mathbb{R}$.
(To be precise, every set $\{a\}$ with $ a \ne 0$ ...)
A: Take the vectorial space $\mathbb R $,
$(1) $ is a base.
$(2) $ is a base.
$(\pi) $ is a base.
if $x\ne 0$, $(x) $ is a base since the dimension is one.
if the dimension is $N $, then each family $(u_1,u_2,...u_N) $ which is span and free is a base.
