Understanding Eigenspaces and Eigenvectors

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SUMMARY

The discussion focuses on the mathematical concepts of eigenspaces and eigenvectors, specifically addressing the equation A\boldsymbol{x} = \lambda\boldsymbol{x}. It confirms that the vectors \boldsymbol{x} forming the eigenspace are linearly independent eigenvectors, provided that the equation A\boldsymbol{x} - \lambda\boldsymbol{x} = \boldsymbol{0} has a nontrivial solution. The process involves solving the determinant det(A - \lambda I) to find eigenvalues and subsequently solving Ax = \lambda x to determine the corresponding eigenvectors.

PREREQUISITES
  • Linear Algebra concepts, particularly eigenvalues and eigenvectors
  • Understanding of matrix operations and determinants
  • Knowledge of nullspace and linear independence
  • Familiarity with the notation and properties of matrices
NEXT STEPS
  • Study the process of calculating eigenvalues using the characteristic polynomial
  • Learn about the geometric interpretation of eigenspaces in vector spaces
  • Explore applications of eigenvectors in systems of differential equations
  • Investigate the role of eigenvalues in Principal Component Analysis (PCA)
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Students and professionals in mathematics, engineering, and data science who are looking to deepen their understanding of linear transformations and their applications in various fields.

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Homework Statement



Suppose [tex]A\boldsymbol{x} = \lambda\boldsymbol{x}[/tex] and [tex]A\boldsymbol{x} - \lambda\boldsymbol{x} = \boldsymbol{0}[/tex]

Then the [tex]\boldsymbol{x}[/tex] (vectors) that form the eigenspace are the linearly independent set of eigenvectors assuming [tex]A\boldsymbol{x} - \lambda\boldsymbol{x} = \boldsymbol{0}[/tex]
has a nontrivial solution.

The Attempt at a Solution



It's true right? It's just solving the nullspace and then naming the new solutions as eigenvectors.
 
Physics news on Phys.org
yes

solving [itex]det(A-\lambda I)[/itex] will give you the allowable values for lambda (eigenvalues of A)

yes

then for a given [itex]\lambda[/itex], solving [itex]Ax = \lambda x[/itex] for non-trivial x, will give you the corresponding eigenvectors for that eigenvalue
 

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