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.


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.

  • $\begingroup$ $\lambda f(1)= \lambda a_0+\lambda a_1 1^1\dots+\lambda a_n1^n =\lambda(a_o+a_1+\dots+a_n)=\lambda\cdot 0=0\in U_1$ ? $\endgroup$ – Doesbaddel May 16 at 8:15
  • 1
    $\begingroup$ @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. $\endgroup$ – Theo Bendit May 16 at 8:46
  • $\begingroup$ Alright, thanks for clarification! $\endgroup$ – 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.


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