How Do You Simplify This Vector Expression?

  • Thread starter Thread starter izchief360
  • Start date Start date
  • Tags Tags
    Vector
izchief360
Messages
7
Reaction score
0
I'm trying to simplify the following expression:

(4u + 3v) ⋅ (4u − 2v) − ll 3u − 4v ll2

And I'm unsure how to proceed.
 
Physics news on Phys.org
Assuming the dot represents the inner product, what formulas do you know that give the properties of the inner product?
 
Is the inner product equivalent to the dot product? The only relevant formula I know is the that of the dot product, but I am unsure of how to apply order of operations when dealing with vectors.
 
Yes, the inner product is the dot product (in the case of real spaces). Dot product is a multiplication, so it has a higher order then addition/subtraction, and the same order as multiplication by a scalar.
 
Thanks folks, I solved it. The process included taking the inside terms of the entire first term and dotting them with the entire second term as follows:
(4u + 3v) ⋅ (4u − 2v)
[4u ⋅ (4u − 2v)] + [3v ⋅ (4u − 2v)]
16u2 - 8u⋅v + 12u⋅v - 6v2

and for the second part, ll 3u − 4v ll2 is equivalent to (3u - 4v)⋅(3u - 4v), and it's the same process as above. Then, just combine like terms.
 
##\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