# Decide whether the following subsets of a vector space are sub-vector spaces.

a) $$\{f\in\mathbb{R}[t]:f(1)=0\}=:U_1$$
b) $$\{f\in\mathbb{R}[t]:\exists a\in\mathbb{R}\text{ with }f(a)=0\}=:U_2$$

where $$\mathbb{R}[t]$$ is the set of all polynomials above K.

Does a) mean, that there is $$f=a_0x^0+a_1x^1+\dots+a_nx^n$$ with $$0=a_0+a_1+\dots+a_n$$ (Basically the zero polynomial)?

b) The same, but $$f=a_0'x^0+a_1'x^1+\dots+a_n'x^n$$ with $$0=a_0'a^0+a_1'a^1+\dots+a'_na^n$$?

If I need to decide whether the following subsets of a vector space are sub-vector spaces, are the following proofs correct?

a) Let $$u,v\in U_1 \implies u=\sum_{i=0}^na_i=0,\;v=\sum_{i=0}^na'_i=0$$
$$\implies 0+0=0\in U_1$$
$$\implies \lambda0=0\in U_1,\quad \lambda\in\mathbb{K}$$ $$\implies U_1$$ is a subvectorspace

b) Not sure how to do this.

## 2 Answers

a) No, it's not correct. You need to remember what these elements of $$U_1$$ are. They are polynomials, which are particular functions. They are not all necessarily of fixed degree $$n$$ either.

Two arbitrary elements $$f, g \in \Bbb{R}[t]$$ take the form \begin{align*} f(t) &= a_0 + a_1 t + a_2 t^2 + \ldots + a_n t^n \\ g(t) &= b_0 + b_1 t + b_2 t^2 + \ldots + b_m t^m, \end{align*} where $$a_0 + a_1 + \ldots + a_n = 0$$ and $$b_0 + b_1 + \ldots + b_m = 0$$. Note that $$n$$ and $$m$$ need not be equal, and $$f$$ is not the same thing as $$a_0 + a_1 + \ldots + a_n$$; this is just one value that $$f$$ takes, when $$t = 1$$, and it happens to be $$0$$.

Adding $$f$$ and $$g$$ gives you $$(f + g)(t) = a_0 + a_1 t + \ldots + a_n t^n + b_0 + b_1 t + \ldots + b_mt^m.$$ Is this polynomial in $$U_1$$? Let's try plugging in $$t = 1$$: \begin{align*} (f + g)(1) &= a_0 + a_1 \cdot 1 + \ldots + a_n \cdot 1^n + b_0 + b_1 \cdot 1 + \ldots + b_m \cdot 1^m \\ &= (a_0 + a_1 + \ldots + a_n) + (b_0 + b_1 + \ldots + b_m) \\ &= 0 + 0 = 0. \end{align*} That is, $$(f + g)(1) = 0$$, so $$f + g \in U_1$$.

Now you can try something similar for scalar multiplication.

b) This is not a subspace. You should be able to find two specific polynomials (with specific coefficients; no arbitrary coefficients $$a_i$$s) with roots, but whose sum has no roots. Try summing a parabola to a linear polynomial.

• $\lambda f(1)= \lambda a_0+\lambda a_1 1^1\dots+\lambda a_n1^n =\lambda(a_o+a_1+\dots+a_n)=\lambda\cdot 0=0\in U_1$ ? – Doesbaddel May 16 at 8:15
• @Doesbaddel That's right, except for the $0 \in U_1$. Remember, it's $\lambda f$, the function, that belongs to $U_1$. The scalar $0$ does not. – Theo Bendit May 16 at 8:46
• Alright, thanks for clarification! – Doesbaddel May 16 at 8:51

The second set is not a vector space. Consider the two polynomials $$x^2$$ and $$\left( x + 1 \right)^2$$. These two are in the set since they have roots at $$0$$ and $$-1$$ respectively. However, their sum, $$x^2 + \left( x + 1 \right)^2$$ can never be zero since one of the two terms is always positive.