Analyzing perturbation of vectors

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SUMMARY

The discussion centers on the mathematical problem involving a non-singular matrix A and a vector b, where the goal is to demonstrate that if Ax = b, then A(x + δx) = b + δb. The user successfully completed the first part of the problem using the definition of non-singularity but encountered difficulties with the second part, specifically regarding the application of linearity and backward stability concepts. Clarification on these mathematical principles is sought to resolve the confusion.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically non-singular matrices.
  • Familiarity with vector perturbation and stability analysis.
  • Knowledge of linear transformations and their properties.
  • Basic proficiency in mathematical proofs and definitions.
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  • Study the properties of non-singular matrices in linear algebra.
  • Learn about perturbation theory and its applications in numerical analysis.
  • Research linearity in transformations and its implications in vector spaces.
  • Explore backward stability and its relevance in solving linear equations.
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Students and professionals in mathematics, particularly those studying linear algebra, numerical analysis, or anyone involved in solving systems of linear equations.

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Homework Statement



We have A \in R^{mxm} \text{ and } b \in R^{m} \text{ and } b \neq 0 \text{. Show that } Ax = b \text{ and } A(x+ \delta x) = b+ \delta b


The Attempt at a Solution



I did the first part just by the definition of A being non singular. The second part is tripping me up though. Not sure how to tackle it. I looked at the backwards stability but it just confused me more. Any help would be most appreciated.
 
Last edited:
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If ##Ax=b## doesn't ##A(x+\delta x)= b +\delta b## just follow by linearity?
 

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