MHB 290 Expanding this determinant about the the second column....

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
View attachment 8905
where does 32 (red) come from ?
nevermind looks its (-8)(4)=-32

but will probable have more ? on this example
 

Attachments

  • 2.png
    2.png
    16.3 KB · Views: 151
Last edited:
Physics news on Phys.org
Expanding on the second column we have
$\left|\begin{array}{ccc}\lambda- 5 & 0 & 4 \\ -12 & \lambda- 1 & 12 \\ -8 & 0 & \lambda- 7\end{array}\right|$$= -(\lambda- 1)\left|\begin{array}{cc} \lambda- 5 & 4 \\ -8 & \lambda+ 7 \end{array}\right|$$= -(\lambda- 1)[(\lambda- 5)(\lambda+ 7)- (-8)(4)]$$=-(\lambda- 1)[(\lambda- 5)(\lambda+ 7)+ 32]$.
The "32" comes from the -8 and 4 in the original matrix.
 
Last edited:
thank you for helping
I am getting the impression not to many here can help with Linear Algebra
despite all the views of curiosity :cool:I did this with another problem but was kinda :confused: with it

https://www.physicsforums.com/attachments/8909
 
Last edited:
##\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...

Similar threads

Replies
14
Views
3K
Replies
10
Views
3K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
4
Views
4K
Back
Top