M^5v given that M = [3 2 4 2 0 2 4 2 3]

  • Context: Undergrad 
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Discussion Overview

The discussion revolves around the calculation of M^5v, where M is a given matrix and v is a column vector. Participants explore the implications of v being an eigenvector of M and how that affects the computation of M raised to a power acting on v.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant inquires about the calculation of M^5v given the matrix M and vector v.
  • Another participant suggests that v is an eigenvector of M and proposes finding the corresponding eigenvalue.
  • A different participant explains the relationship between M, its eigenvalues, and powers of M acting on v, indicating that M^2v can be expressed in terms of the eigenvalue.
  • Another contribution emphasizes that M^5v can be simplified due to the properties of eigenvectors, suggesting that repeated application of M can be streamlined.

Areas of Agreement / Disagreement

Participants appear to agree on the notion that v is an eigenvector of M and that this property simplifies the calculation of M^5v. However, the specific eigenvalue and further implications remain unaddressed, indicating that the discussion is not fully resolved.

Contextual Notes

The discussion does not clarify the specific eigenvalue associated with v or the implications of this eigenvalue on the calculation of M^5v. There may be assumptions about the properties of eigenvectors that are not explicitly stated.

Who May Find This Useful

This discussion may be useful for individuals interested in linear algebra, particularly those studying eigenvalues and eigenvectors, as well as their applications in matrix operations.

squenshl
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Hi,
Just wondering how to calculate M^5v given that
M = [3 2 4
2 0 2
4 2 3].
and v = (1,0,-1). v being a column vector.
 
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v is an eigenvector of M. Find the eigenvalue that it corresponds to.
 


If [itex]Mv= \lambda v[/itex], then [itex]M^2v= M(Mv)= M(\lambda v)= \lambda Mv= \lambda^2 v[/itex]. Get the point?
 


Also, remember that M^5 is just M acting on v 5 consecutive times. Eigenvectors make this quite simple :)
 

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