1. ## Polynomial vector spaces

Hi,

problem:
Consider the vector space $\displaystyle \mathbb{P}$ and the subsets $\displaystyle \mathbb{V}\;of\;\mathbb{P}$ consisting of those vectors (polynomials) $\displaystyle x$ for which

(a) x has degree 3,

(b) 2x(0)=x(1)

(c) $\displaystyle x(t)\geq0$ whenever $\displaystyle 0\leq t\leq 1$,

(d) $\displaystyle x(t)=x(1-t)$ for all $\displaystyle t$.

In which of these cases is $\displaystyle \mathbb{V}$ a vector space?

attempt:
I first assume that $\displaystyle \mathbb{P}$ is a vector space over $\displaystyle \mathbb{C}$ as this is not mentioned in the problem.

(a) Not a vector space.
$\displaystyle x(t)=a_3t^3+a_2t^2+a_1t+a_0$
Closed under scalar multiplication.
It does not have an origin that is a polynomial of degree 3 such that $\displaystyle x(t)+0(t)=x(t)$.

(b) Is a vector space.
$\displaystyle x(t) = a_nt^n+\cdots+a_1t+(a_n+\cdots+a_1)$
Closed under scalar multiplication.
There exists an element (origin) $\displaystyle 0(t)$, such that $\displaystyle x(t)+0(t)=x(t)$
$\displaystyle 2\cdot0(0)=0(1)$

(c) and (d), could need a push in the right direction here.

Thanks!

2. Originally Posted by Mollier
Hi,

problem:
Consider the vector space $\displaystyle \mathbb{P}$ and the subsets $\displaystyle \mathbb{V}\;of\;\mathbb{P}$ consisting of those vectors (polynomials) $\displaystyle x$ for which

(a) x has degree 3,

(b) 2x(0)=x(1)

(c) $\displaystyle x(t)\geq0$ whenever $\displaystyle 0\leq t\leq 1$,

(d) $\displaystyle x(t)=x(1-t)$ for all $\displaystyle t$.

In which of these cases is $\displaystyle \mathbb{V}$ a vector space?

attempt:
I first assume that $\displaystyle \mathbb{P}$ is a vector space over $\displaystyle \mathbb{C}$ as this is not mentioned in the problem.

(a) Not a vector space.
$\displaystyle x(t)=a_3t^3+a_2t^2+a_1t+a_0$
Closed under scalar multiplication.
It does not have an origin that is a polynomial of degree 3 such that $\displaystyle x(t)+0(t)=x(t)$.

(b) Is a vector space.
$\displaystyle x(t) = a_nt^n+\cdots+a_1t+(a_n+\cdots+a_1)$
Closed under scalar multiplication.
There exists an element (origin) $\displaystyle 0(t)$, such that $\displaystyle x(t)+0(t)=x(t)$
$\displaystyle 2\cdot0(0)=0(1)$

(c) and (d), could need a push in the right direction here.

Thanks!

(c) Take an element here and multiply it by the scalar -1

(d) This is a subspace, and you've to check it directly.

Tonio

3. Originally Posted by Mollier
Hi,

problem:
Consider the vector space $\displaystyle \mathbb{P}$ and the subsets $\displaystyle \mathbb{V}\;of\;\mathbb{P}$ consisting of those vectors (polynomials) $\displaystyle x$ for which

(a) x has degree 3,

(b) 2x(0)=x(1)

(c) $\displaystyle x(t)\geq0$ whenever $\displaystyle 0\leq t\leq 1$,

(d) $\displaystyle x(t)=x(1-t)$ for all $\displaystyle t$.

In which of these cases is $\displaystyle \mathbb{V}$ a vector space?

attempt:
I first assume that $\displaystyle \mathbb{P}$ is a vector space over $\displaystyle \mathbb{C}$ as this is not mentioned in the problem.

(a) Not a vector space.
$\displaystyle x(t)=a_3t^3+a_2t^2+a_1t+a_0$
No, it is not closed under vector addition.
If $\displaystyle x(t)= -t^3+ 2t^2+ 3t+ 1$ and $\displaystyle y(t)= t^3+ 3t^2- t+ 2$ then $\displaystyle x(t)+ y(t)= 5t^2+ 2t+3$ which is NOT third degree.

Closed under scalar multiplication.
No, it is not closed under scalar multiplication. 0 is a scalar and 0 time any third degree polynomial is NOT a third degree polynomial.

It does not have an origin that is a polynomial of degree 3 such that $\displaystyle x(t)+0(t)=x(t)$.
But of course, that just says this is NOT a vector space- but you could have stopped after showing it is not closed under addition.

(b) Is a vector space.
$\displaystyle x(t) = a_nt^n+\cdots+a_1t+(a_n+\cdots+a_1)$
Closed under scalar multiplication.
You need to show these, not just say they are true. If 2x(0)= x(1), 2y(0)= y(1) and z(t)= x(t)+ y(t), then 2z(0)= 2(x(0)+ y(0))= 2x(0)+ 2y(0)= x(1)+ y(1)= z(1) and, for any scalar, a, 2(az(t))= 2(ax(t)+ ay(t))= a(2x(t))+ a(2y(t))= ax(1)+ ay(1)= a(x(1)+ y(1))= az(1).

There exists an element (origin) $\displaystyle 0(t)$, such that $\displaystyle x(t)+0(t)=x(t)$
$\displaystyle 2\cdot0(0)=0(1)$
Good! You showed that the 0 function is in this set.

(c) and (d), could need a push in the right direction here.
Suppose x(t) is such that $\displaystyle x(t)\ge 0$ for [/tex]0\le t\le 1[/tex]. What about (-1)x?

(d) is pretty much like the others. If x(t)= x(1-t), y(t)= y(1-t), and z= x+ y, then z(t)= x(t)+ y(t)= x(1-t)+ y(1-t)= z(1- t), etc.
Thanks!

4. (c)
$\displaystyle (-1)x(t) \leq 0$, i.e. not closed under scalar multiplication.

(d)
Scalar multiplication:
\displaystyle \begin{aligned} az(t)=ax(t)+ay(t)=&\;a(x(t)+y(t))\\ =&\;a(x(1-t)+y(1-t))\\ =&\;az(1-t) \end{aligned}

Origin:
$\displaystyle z(t)+0(t)=z(t)$

Now I guess I need to show that $\displaystyle 0(t)=0(t-1)$.

\displaystyle \begin{aligned} z(t)+0(t)=&\;z(t-1)+0(t-1)\\ z(t)-z(t-1)+0(t)=&\;z(t-1)-z(t-1)+0(t-1)\\ 0(t)=&\;0(t-1) \end{aligned}

Again, thank you very much for dumbing down replies for me HallsOfIvy
I dream of the day you won't have to!