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?

  • $\begingroup$ Proper subspace means a subspace that is not $\Bbb R[x]$. $\endgroup$
    – Arnaud D.
    May 11, 2017 at 10:20
  • 2
    $\begingroup$ Hint: consider $p(x) \mapsto x p(x)$. $\endgroup$
    – Kenny Wong
    May 11, 2017 at 10:28
  • 1
    $\begingroup$ If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, \ldots)$ where each $a_i \in R$. There is nothing polynomial-specific being used here. $\endgroup$
    – Joppy
    May 11, 2017 at 13:26

2 Answers 2


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.


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.

  • $\begingroup$ Can you show a bit more hint I am very new to this topic. Thank you! $\endgroup$
    – stedmoaoa
    May 11, 2017 at 12:13
  • $\begingroup$ So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to? $\endgroup$
    – stedmoaoa
    May 25, 2017 at 10:18
  • $\begingroup$ I got the idea but how to show that it is one-to-one and onto? $\endgroup$
    – stedmoaoa
    May 25, 2017 at 10:20
  • $\begingroup$ @stedmoaoa, just use the function definition I gave. $\endgroup$
    – lhf
    May 25, 2017 at 11:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .