Can we find Eigenvalues for simultaneous equation?

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Discussion Overview

The discussion revolves around the limitations and conditions for finding eigenvalues, particularly in the context of simultaneous equations and the implications of the fundamental theorem of algebra. The scope includes theoretical aspects of linear algebra and eigenvalue problems.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • One participant inquires about the limitations for finding eigenvalues.
  • Another participant states that, according to the fundamental theorem of algebra, there is always at least one complex solution to the characteristic polynomial, implying at least one complex eigenvalue exists.
  • A further contribution reiterates the existence of at least one complex eigenvalue and mentions that every finite-dimensional real vector space has an invariant subspace of degree one or two.
  • It is noted that the existence of eigenvalues is true only in the finite-dimensional case.

Areas of Agreement / Disagreement

Participants generally agree that there is always at least one complex eigenvalue in finite-dimensional spaces, but the discussion does not resolve the initial inquiry about limitations or conditions for finding eigenvalues.

Contextual Notes

The discussion does not address specific limitations or assumptions regarding the types of matrices or conditions under which eigenvalues may be found.

wasi-uz-zaman
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hi, please tell me what are the limitations for finding eigenvalues ?
thanks
 
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Could you be more specific? By the fundamental theorem of algebra, there always exists at least one (complex) solution to the characteristic polynomial so there will always be at least one (complex) eigenvalue, where complex means a number of the form ##a + bi##.
 
WannabeNewton said:
Could you be more specific? By the fundamental theorem of algebra, there always exists at least one (complex) solution to the characteristic polynomial so there will always be at least one (complex) eigenvalue, where complex means a number of the form ##a + bi##.
Indubitably.

And every finite dimensional real vector space has an invariant subspace of degree one OR two.
 
WannabeNewton said:
Could you be more specific? By the fundamental theorem of algebra, there always exists at least one (complex) solution to the characteristic polynomial so there will always be at least one (complex) eigenvalue, where complex means a number of the form ##a + bi##.

This is of course only true in the finite-dimensional case.
 

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