MHB Finding a 3rd polynomial to create a basis.

  • Thread starter Thread starter Harambe1
  • Start date Start date
  • Tags Tags
    Basis Polynomial
Harambe1
Messages
5
Reaction score
0
Hi,

I am struggling with the following problem:

"Let $V=P_3(\Bbb{R})$ and let $t_1=3x^3-x-2$ and $t_2=x^3-3x+2$ with $T=\left\{ t\in V \:|\: t(1)=0 \right\}$. Find ${t_3}\in\left\{T\right\}$ such that $\left\{t_1, t_2, t_2\right\}$ is a basis of T.

Not sure where to go as each column matrix will have 4 elements but then there is only 3 polynomials in the basis.Thanks.
 
Physics news on Phys.org
Harambe said:
Hi,

I am struggling with the following problem:

"Let $V=P_3(\Bbb{R})$ and let $t_1=3x^3-x-2$ and $t_2=x^3-3x+2$ with $T=\left\{ t\in V \:|\: t(1)=0 \right\}$. Find ${t_3}\in\left\{T\right\}$ such that $\left\{t_1, t_2, t_2\right\}$ is a basis of T.

Not sure where to go as each column matrix will have 4 elements but then there is only 3 polynomials in the basis.Thanks.

Hi Harambe! Welcome to MHB! ;)

Indeed, there are 4 elements, so a basis of $V$ will have 4 polynomials.
However, $T$ has the restriction $t(1)=0$, meaning we will be left with a basis of 3 polynomials.

Do you have any ideas how we might find that 3rd polynomial?
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top