Basis Represented By Standard Basis Is it true that all basis vectors can be represented by the standard basis?

 for example every basis in $R^2$ like $(-2,1),(1,1)$ are linear combination of  $(1,0),(0,1)$
 A: The definition of a basis of a vector space means it is a family $\def\B{\mathcal B}\B$ of vectors such that any (single) vector$~v$ can be expressed uniquely as linear combination of the vectors in$~\B$. The coefficients used in that linear combination are called the coordinates of $v$ (with respect to$~\B$). Note that any linear combination gives you one vector, not another basis. So expressing all vectors of a another basis with respect to$~\B$ gives a square matrix of coefficients (a column for each vector if the new basis) which is called a change of basis matrix.
A standard basis only exists in vector spaces $\mathbf R^n$ (and not for instance in subspaces of it, nor in spaces where vectors are different things to begin with, such as functions). Since here vectors are $n$-tuples of numbers, the standard basis has the property that every vector is equal to its collection of coordinates with respect to the basis. In other words expressing a vector in the standard basis requires no work at all, the required coefficients can be taken from the guts of the vector itself.
So to answer the question, in $\mathbf R^2$ (and more generally in $\mathbf R^n$) all vectors can be trivially expressed in the standard basis, and if you've got another basis, then this applies in particular to its vectors (the term "basis vector" does not really have a sense in isolation, as you use it). But "all vectors can be represented by the standard basis" is not true in general, because there is not always such a thing as a standard basis.
A: Just see this

$$ (-2,1) = -2(1,0)+(0,1) $$

which is a linear conmination of the standard basis.
