Finding the eigenvectors (and behavior of solution) around the

  • Context: Graduate 
  • Thread starter Thread starter Somefantastik
  • Start date Start date
  • Tags Tags
    Behavior Eigenvectors
Click For Summary
SUMMARY

This discussion focuses on finding the eigenvectors associated with the critical points of the matrix D_{f} = \[\begin{pmatrix}32x & 18y \\ 32x & -32y\end{pmatrix}\]. The specific evaluation at the point (1,1) yields D_{f}(1,1) = \[\begin{pmatrix}32 & 18 \\ 32 & -32\end{pmatrix}\], leading to the characteristic polynomial det(A-\lambda I) = -256 + \lambda^{2} - 146, which results in eigenvalues λ_{1} = 20 and λ_{2} = -20. The first eigenvector is determined as ξ^{(1)} = \left[^{9}_{4} \right], while the second eigenvector ξ^{(2)} is initially miscalculated as [0 0]T, but the issue was resolved during the discussion.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors
  • Familiarity with matrix operations and determinants
  • Knowledge of LaTeX for mathematical notation
  • Basic concepts of linear algebra
NEXT STEPS
  • Study the process of calculating eigenvectors for 2x2 matrices
  • Learn about the implications of eigenvalues in stability analysis
  • Explore advanced matrix decomposition techniques
  • Review the use of LaTeX for presenting mathematical equations
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are working with linear algebra concepts, particularly those focusing on eigenvalues and eigenvectors in dynamic systems.

Somefantastik
Messages
226
Reaction score
0
finding the eigenvectors (and behavior of solution) around the critical points found in this thread: https://www.physicsforums.com/showthread.php?t=258349&referrerid=110346

[tex]D_{f} = \[\begin{pmatrix}32x & 18y \\ 32x & -32y\end{pmatrix}\][/tex]

[tex]D_{f}(1,1) = \[\begin{pmatrix}32 & 18 \\ 32 & -32\end{pmatrix}\][/tex]

[tex]= \[\begin{pmatrix}16 & 9 \\ 16 & -16 \end{pmatrix}\][/tex]

[tex]det(A-\lambda I) =\[\begin{pmatrix} 16-\lambda & 9 \\ 16 & -16- \lambda \end{pmatrix}\][/tex]

[tex]= -256 + \lambda^{2} - 146 \ => \ \lambda = ^{+}_{-}20[/tex]

[tex]\lambda_{1} = 20:[/tex]

[tex](A-\lambda_{1} I)\xi^{(1)} = 0 \ => \ \[\begin{pmatrix} -4 & 9 \\ 16 & -36 \end{pmatrix}\]\xi^{(1)} = 0[/tex]

I can't get LaTeX to cooperate with me, that's supposed to say [-4 9; 16 -36]ξ(1) = 0

[tex]=> \ \xi^{(1)} = \left[^{9}_{4} \right][/tex]

Having trouble finding [tex]\xi^{(2)}[/tex] when [tex]\lambda_{2} = -20[/tex].

Keeps coming out to be [0 0]T.

Any suggestions?
 
Physics news on Phys.org


Looks like you can't reduce the matrix before you do det(A - tI). Figured it out; thanks for looking.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
9
Views
2K