# Prove if $W \subset V$ is a subspace such that $T(W)\subset W$, then $V/W$ is finite-dimensional.

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.

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

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.

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 $\partial (v-w) < n$.