Lin. Algebra - Find Eigenvectors / eigenvalues

  • #1
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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) = [tex]\lambda[/tex](z1, z2, z3).

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

[tex]\lambda[/tex]z1 = 2*z2
[tex]\lambda[/tex]z2 = 0
[tex]\lambda[/tex]z3 = 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:

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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From the second equation, either [itex]\lambda= 0[/itex] or z2= 0. If [itex]\lambda[/itex] is not 0, the z2= 0, and so z1= 0. The last equation is [itex]\lambda[/itex]z3= 5z3 so either z3= 0 or [itex]\lambda= 5[/itex]. Yes [itex]\lambda[/itex]= 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 [itex]\lambda[/itex]= 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.
 
  • #3
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Thanks for making things more clear. That helped a lot.
 

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