Existence of Standard Basis Do all vector spaces have standard bases? 
I was reading a book which says that "it makes no sense talking about standard bases for any vector space". I don't really understand this concept. 
If this is right, under what conditions does a vector space have a standard basis?
 A: A vector space has a standard basis if and only if people have selected one particular basis and given it the name "standard". In the case of $\mathbb R^n$, people have done that, with the basis consisting of vectors with a single component equal to $1$ and all other components $0$. Furthermore, this particular choice has been so widely accepted that you can say "standard basis for $\mathbb R^n$ and reasonably expect all mathematicians to understand what you mean. 
For other vector spaces, someone might choose a basis and call it standard, especially if this happens in the context of a paper in which that basis is particularly useful. But there might be no general agreement among mathematicians that this particular basis deserves the name "standard". So it might be called "the standard basis" in a particular paper, but not in other contexts.  (Furthermore, different people, working in different contexts, might have proposed different bases as "standard" for the same space.)
There are also plenty of vector spaces for which no one has (yet) selected a particular basis and named it "standard".  
I would use the phrase "standard basis" only in two situations: (1) Cases like $\mathbb R^n$ where we all agree as to what the standard basis is, so I could refer to it without fear of confusion. (2) Cases where I've chosen a particular basis, decided to call it "standard", and explicitly said (earlier in the same paper) which basis I meant by "standard basis". 
A: Consider the vector space consisting out of real polynomials of degree less than or equal to $n-1$. That is:
$$P_{n-1} = \{a_{n-1}x^{n-1} + \ldots + a_1x + a_0\}.$$
A basis for this vectorspace is given by 
$$\{1, x, x^2, \ldots, x^{n-1}\}$$
(check this). Now any polynomial of degree less than or equal to $n-1$ is uniquely determined by the coefficients with respect to this basis: for example the polynomial $x^2 + x -1$ in $P_3$ has coefficients $(-1, 1, 1, 0)$. (Do you see this?) But we can also describe the basis elements in coordinates with respect to this basis: for example
$$x = 0 \cdot 1 + 1 \cdot x + 0 \cdot x^2 + 0 \cdot x^3$$
in the case of $P_3$. written as a vector: $(0,1,0,0)$. 
This is what we need to when we want to describe the matrix corresponding to some linear transformation: suppose $V$ and $W$ are vectorspaces of dimension $n$ respectively $m$. We want to describe the matrix corresponding to some linear transformation $L: V \to W: v \mapsto L(v)$. In order to be able to do this, we need to have bases of $V$ and $W$. This allows us to describe all elements of $V$ respectively $W$ using their coordinates with respect to the chosen basis of $V$ respectively the chosen basis of $W$.After we have choosen this basis, we find that the coordinates of the basisvectors with respect to the basisvector are exactly the vectors $(1, 0, \ldots, 0)$ etc. I hope this solves your problem lot.
NOTE: although this does not seem to answer the original question of the OP, I have posted this answer in the light of our discussion in the comments on this question. This answer however was way too long to post as a comment.
