Show that the vector space of polynomials R[x] is isomorphic to a proper subspace of itself Show that the vector space of polynomials R[x] is isomorphic to a proper
subspace of itself:
Vector Space Isomorphism exists when there exists a bijective (one-to-one and onto) linear mapping F:V $\rightarrow$U. 
the coefficient of the polynomials can be written as $(a_0,a_1,a_2...)$. But how to find the subspace?
What about instead to prove the dimension of the the two vector spaces is the same, which means isomorphic? But how to do it?
 A: First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $n\ge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
A: Hint: Consider $p(x) \mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them:
$$(a_0,a_1,a_2,\dots,a_n,0,0,0,\dots) \mapsto (a_0,0,a_1,0,a_2,\dots,a_{n-1},0,a_n,0,0,0,\dots)$$
 and so you can recover one from the other.
