MHB 051 how they got the eigenspaces ?

  • Thread starter Thread starter karush
  • Start date Start date
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
051.png


ok I didn't understand how they got the eigenspaces

the original matrix was
$A=\left[\begin{array}{rrr}−1&2\\−6&6\end{array} \right]$
so think I got values correct $\lambda=2,3$

https://dl.orangedox.com/wlKD7eKSWiQ79alYD6
 
Last edited:
Physics news on Phys.org
Yes, the eigenvalues are 3 and 4.

Any eigenvector corresponding to eigenvalue 3 must satisfy
$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}3x \\ 3y \end{bmatrix}$
so -x+ 2y= 3x and -6x+6y= 3y. What must x and y be? (Those two equations reduce to the same thing so there are an infinite number of solutions- a one dimensional subspace.)

Any eigenvector corresponding to eigenvalue 4 must satisfy
$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}4x \\ 4y \end{bmatrix}$
so -x+ 2y= 4x and -6x+6y= 4y. What must x and y be?

Those eigenvectors span the "eigenspace".
 
Last edited:
Country Boy said:
Any eigenvector corresponding to eigenvalue 4 must satisfy
$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}4x \\ 4y \end{bmatrix}$
so -x+ 2y= 4x and -6x+6y= 4y. What must x and y be?

Those eigenvectors span the "eigenspace".
That last is wrong because, of course, the eigenvalue was 2, not 4!
Instead -x+ 2y= 2x and -6x+ 6y= 2y.
2y= 3x and -6x= -4y both reduce to 3x= 2y. In particular, if we take x= 2, then 3x= 6= 2y, y= 3. One eigenvector is (2, 3) but any multiple is also an eigenvector.
 
##\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
2
Views
3K
Replies
3
Views
1K
Replies
7
Views
1K
Replies
2
Views
973
Replies
3
Views
1K
Replies
14
Views
2K
Back
Top