1
$\begingroup$

Let $V$ denote the vector space of polynomials in one variable with coefficients in $\mathbb{R}$, and let $$T(f(x)) = xf(x).$$ Prove that if $W \subset V$ is a subspace such that $T(W)\subset W$, then $V/W$ is finite-dimensional.

First I let $f(x) \in W,$ such that $a_0+a_1x+a_2x^2+...+a_nx^n$, and $T(f(x)) = a_0x+a_1x^2+a_2x^3+...+a_nx^{n+1} \in W$. So I think $W$ is the set contains polynomials such that $a+ax+ax^2+ ... +ax^n +...ax^\infty$. But I am not sure about it, and how it is related to finite-dimension.

Please help me to solve this problem, thank you!!

$\endgroup$
  • 1
    $\begingroup$ Hint : consider the minimal degree of a non-zero polynomial in $W$. $\endgroup$ – Captain Lama Mar 29 '16 at 19:42
  • 3
    $\begingroup$ What if $W= \{0\}$? $\endgroup$ – copper.hat Mar 29 '16 at 19:44
3
$\begingroup$

You need to show the following claim:

For all $f(x) \in V$, for all $g(x) \in W$, necessarily $f(x)g(x) \in W$.

To show this, write $f(x)= a_n x^n + \dots +a_1x + a_0$. Then $$a_0g(x) \in W$$ $$a_1g(x) \in W \Rightarrow a_1xg(x) \in W$$ $$\vdots$$ $$a_ng(x) \in W \Rightarrow a_nxg(x) \in W \Rightarrow \dots \Rightarrow a_nx^ng(x) \in W$$ hence the sum $$a_0 g(x) + a_1xg(x) + \dots + a_nx^ng(x) = (a_0+a_1x+ \dots +a_n x^n )g(x) = f(x)g(x) \in W$$ and the claim is proved.

From now on, we need the assumption $W \neq 0$.

Now, if you know some of ring theory, you should recognize that this means that $W$ is an ideal of $V= \Bbb{R}[x]$, hence $W=\langle m(x) \rangle$ for some generating polynomial $m(x)$, and $V/W$ has dimension $\deg m(x)$.

Otherwise, one can directly show that

$V/W$ is generated by $$1+W , x+W , \dots , x^{d-1}+W$$ hence $\dim (V/W) \le d$.

In fact, for all $f(x) + W \in V/W$, divide $$f(x) = q(x)m(x) + r(x)$$ where $q(x)$ is the quotient, and $r(x)$ is the remainder of the division, with $\deg r(x) < d$ or $r(x)=0$. Then $$f(x)+W = q(x)m(x)+r(x)+W = r(x)+W$$ and you conclude.

$\endgroup$
1
$\begingroup$

Here is a non ideal approach:

Hint 1: If $W$ contains at least one non zero point, then it must contain polynomials of arbitrarily large degree.

Hint 2: There is some $n$ such that $W$ contains polynomials of degree $m$ for any $m \ge n$ (this follows from the definition of $T$).

Hint 3: If $v \in V$ and $\partial v \ge n$, then there is some $w \in W$ such that $\partial (v-w) < n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.