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Thread: Show that λ is an eigenvalue of the B matrix iff λ is an eigenvalue of B'

  1. #1

    Show that λ is an eigenvalue of the B matrix iff λ is an eigenvalue of B'

    Show that λ is an eigenvalue of the B matrix if, and only if λ is an eigenvalue of B'

    It's supposed to be really easy to show that but I guess that it is only the case when you know how to do it :P

  2. #2
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    Let's be clear. Please state what B' is. Notation isn't always uniform.

    Is [tex]|\lambda I - B|[/tex] any different from [tex]|\lambda I - B'|[/tex]
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Quote Originally Posted by tkhunny View Post
    Let's be clear. Please state what B' is. Notation isn't always uniform.

    Is [tex]|\lambda I - B|[/tex] any different from [tex]|\lambda I - B'|[/tex]
    B is the transposed matrix of B!

  4. #4
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    Fair enough. Now the answer to my determinant question...
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
    Quote Originally Posted by tkhunny View Post
    Fair enough. Now the answer to my determinant question...
    Oh sorry! I don't know the only thing I have is the question I wrote :/

    How would it affect the answer?

  6. #6
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    Pick any 2x2 matrix and play with it a bit.
    Expand your exploration from there.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  7. #7
    Quote Originally Posted by tkhunny View Post
    Pick any 2x2 matrix and play with it a bit.
    Expand your exploration from there.
    So I tried with the matrix B (first row 1,2 ; second row 3,4) so B' (first row 1,3 ; second row 2,4) and it gives me | λ I - B | = | λ I - B' |


    But I'm still really not sure how to resolve my problem!

  8. #8
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    You have answered my question for this single matrix. What does it mean for the Eigenvalues if you get EXACTLY the same Characteristic equation?

    Does your example extend to an arbitrary square matrix?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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