True or false eigenvalue problem

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SUMMARY

The statement that different eigenvectors corresponding to an eigenvalue of a matrix must be linearly dependent is false. Eigenvectors corresponding to distinct eigenvalues of a linear operator are linearly independent. The discussion illustrates this by showing that if two eigenvectors, u and v, correspond to distinct eigenvalues λ1 and λ2, assuming they are dependent leads to a contradiction, confirming their independence. This conclusion holds true under the condition that the eigenvalues are distinct and non-zero.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically eigenvalues and eigenvectors.
  • Familiarity with linear operators and their properties.
  • Knowledge of linear independence and dependence in vector spaces.
  • Basic skills in matrix operations and calculations.
NEXT STEPS
  • Study the properties of eigenvalues and eigenvectors in detail.
  • Learn about the implications of distinct eigenvalues on the linear independence of eigenvectors.
  • Explore examples of matrices and calculate their eigenvectors to reinforce understanding.
  • Investigate the concept of diagonalization and its relation to eigenvalues and eigenvectors.
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Students and professionals in mathematics, particularly those studying linear algebra, as well as researchers and educators looking to deepen their understanding of eigenvalue problems and their applications.

achuthan1988
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Q)Different eigenvectors corresponding to an eigenvalue of a matrix must be linearly dependent?
Is the above statement true or false.Give reasons.
 
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achuthan1988 said:
Q)Different eigenvectors corresponding to an eigenvalue of a matrix must be linearly dependent?
Is the above statement true or false.Give reasons.

What did you try already?? If we know what you tried then we'll know where to help??

Begin by looking at some examples of matrices. Pick some arbitrary (easy) matrices and calculate its eigenvectors.
 
I would NOT look at specific matrices. This can be done better more abstractly.

Suppose u and v are non-zero eigenvectors, for linear operator A, corresponding to distinct eigenvalues [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] respectively. Then [itex]Au= \lambda_1 u[/itex] and [itex]Av= \lambda_2 v[/itex].

Suppose they are not independent. Then [itex]v= \mu u[/itex] for some non-zero scalar [itex]\mu[/itex]. That gives [itex]Av= \mu Au[/itex] or [itex]\lambda_1 v= \mu \lambda_2 v[/itex] so that [itex]v= (\mu \lambda_2)/\lambda_1)u[/itex].

But that means [itex](\mu \lambda_2)/\lambda_1= \mu[/itex] so that [itex]\lambda_2= \lambda_1[/itex], a contradiction.

(Of course, this requires [itex]\lambda_1\ne 0[/itex] so we can divide by it. If [itex]\lambda_1= 0[/itex], just reverse the [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex]. They cannot both be 0 because they are distinct.)

Now, can you extend that to any number of independent eigenvectors?
 

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