Lin. Algebra - Find Eigenvectors / eigenvalues

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SUMMARY

The discussion focuses on finding eigenvalues and eigenvectors for the linear transformation T defined in L(F3) by T(z1, z2, z3) = (2*z2, 0, 5*z3). The eigenvalues identified are 5 and 0, with corresponding eigenvectors (0, 0, 1) for the eigenvalue 5 and any multiple of (1, 0, 0) for the eigenvalue 0. The solution process involves setting up a system of equations derived from the transformation and solving for the eigenvalues and eigenvectors.

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  • Understanding of linear transformations in vector spaces
  • Familiarity with eigenvalues and eigenvectors concepts
  • Knowledge of solving systems of linear equations
  • Basic proficiency in linear algebra notation and terminology
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  • Study the properties of eigenvalues and eigenvectors in linear algebra
  • Learn about diagonalization of matrices and its applications
  • Explore the characteristic polynomial and its role in finding eigenvalues
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Students studying linear algebra, educators teaching eigenvalue problems, and anyone seeking to deepen their understanding of linear transformations and their properties.

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Homework Statement


Define T in L(F3) by T(z1, z2, z3) = (2*z2, 0, 5*z3). Find all eigenvalues and eigenvectors of T.

Homework Equations


The Attempt at a Solution


Well, since we want to find all the eigenvalues, we want the following equation to hold:
T(z1, z2, z3) = (2*z2, 0, 5*z3) = \lambda(z1, z2, z3).

Setting up a system of equations, I get the following:

\lambdaz1 = 2*z2
\lambdaz2 = 0
\lambdaz3 = 5*z3

Solving, I get one eigenvalue, namely 5. Then I came up with the corresponding eigenvector, (0, 0, 5). Is this correct?

Edit: 0 is the other eigenvalue, but I don't know what the corresponding eigenvector would be.
 
Last edited:
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From the second equation, either \lambda= 0 or z2= 0. If \lambda is not 0, the z2= 0, and so z1= 0. The last equation is \lambdaz3= 5z3 so either z3= 0 or \lambda= 5. Yes \lambda= 5 is an eigenvalue. That gives no restriction on z3 at all. Any multiple of (0, 0, 1) is an eigenvector.

But don't forget the other possibility. If \lambda= 0 then 5z3= 0 so z3= 0. Also 2z2= 0 so z2= 0. But that gives no restriction on z1! For any z1, T(z1, 0, 0)= (0, 0, 0) so 0 is an eigenvalue with any multiple of (1, 0, 0) as eigenvector.
 
Thanks for making things more clear. That helped a lot.
 

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