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Eigenvalues are invariant but eigenvectors are not

  1. Sep 20, 2015 #1
    Hi there.

    How would I show that the eigenvalues of a matrix are an invariant, that is, that they depend only on the linear function the matrix represents and not on the choice of basis vectors. Show also that the eigenvectors of a matrix are not an invariant.

    Explain why the dependence of the eigenvectors on the particular basis is exactly what we would expect and argue that is some sense they are indeed invariant.

    Do I use the fact that if 2 matrices ##A## and ##B## which are similar, then there exists an invertible matrix ##P## such that ##A = P^{-1}BP##. Hence, there determinants are the same

    Someone please help.
     
  2. jcsd
  3. Sep 20, 2015 #2

    mfb

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    I moved the thread to the homework section.

    The same determinant is not sufficient (different eigenvalues can lead to the same determinant), but you can extend that argument.
     
  4. Sep 20, 2015 #3

    pasmith

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    Hint: [itex]\lambda[/itex] is an eigenvalue of [itex]A[/itex] if and only if [itex]\det(A - \lambda I) = 0[/itex].
     
  5. Sep 20, 2015 #4
    Yes. So then do I show that B has the same determinant hence the same eigenvalues (as B is similar to A)??
     
  6. Sep 20, 2015 #5

    vela

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    Are you saying det(A)=det(B) implies A and B have the same eigenvalues? If so, then no. pasmith is saying to look at det(A-λI), which isn't the same as det(A).
     
  7. Sep 20, 2015 #6
    No I'm saying ##\text{Det}(A-\lambda I) = \text{Det}(B-\lambda I)##. Meaning that A and B have the same eigenvalues.

    I think I'm just not understanding the hint sorry.

    Cheers.
     
  8. Sep 20, 2015 #7

    mfb

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    That is not true.
    You can show that both A and B have the eigenvalue λ if both sides are equal to zero. If they are non-zero, you just know that λ is not an eigenvalue of either matrix.
     
  9. Sep 20, 2015 #8
    Oops right of course I meant to state that ##\text{Det}{A-\lambda I)=0=\text{Det}{A-\lambda I)##.

    So this is sufficient to show eigenvalues are invariant I'm assuming then???

    How do we show eigenvectors are not???
     
  10. Sep 20, 2015 #9

    andrewkirk

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    The eigenvectors do not change. Their representations change. The representation is an n-tuple of numbers.

    The question is better expressed as 'How do I show that the eigenvalues of a linear transformation do not depend on the basis used, but the representation of each eigenvector in a basis depends on the basis'.

    When it's put in that more pedantic way, it is obvious why the representations of the eigenvector change.

    The reason why neither the eigenvalues nor the eigenvectors change is that the eigenvector-eigenvalue pairs ##(\lambda_i,\vec{v}_i)## are the solutions of the equation

    $$L\vec{v}=\lambda\vec{v}$$

    where ##L:V\to V## is a given linear transformation and ##V## is a ##n## dimensional vector space. Since the equation is well-defined and does not specify a basis, its solutions cannot depend on the choice of basis. The eigenvalues and eigenvectors depend only on ##L##, not on ##L## plus a basis.

    Since the ##\lambda## are scalars and so not in the space ##V##, they do not need to be represented in a basis, hence there is no basis representation to vary by basis. On the other hand, the eigenvectors ##\vec{v}## are elements of the vector space ##V##, and hence have a representation in each basis. To show the representation varies by basis, just write the equation for converting the representation of a vector in basis A to a representation in basis B. That involves multiplication by a matrix. Unless that matrix is the identity, the representation will change.
     
  11. Sep 21, 2015 #10

    mfb

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    Careful. I was about to post the same comment, but squenshl is not asking about eigenvectors of a linear transformation. We actually change the matrix, so the eigenvectors of this matrix change.
     
  12. Sep 21, 2015 #11
    Right!!!!

    Nothing to do with linear transformations
     
  13. Sep 21, 2015 #12

    Fredrik

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    No need to involve determinants. Just take the eigenvalue equation ##Ax=\lambda x## and multiply it with something that enables you to rewrite the left-hand side so that it involves B. (This strategy answers both questions).
     
    Last edited: Sep 22, 2015
  14. Sep 21, 2015 #13

    andrewkirk

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    I agree.

    But that assumes that we regard the n-tuple of numbers as the vector, rather than a representation in a basis of a vector, and the matrix as a n x n array of numbers that defines a linear transformation, rather than a representation in a basis of a linear transformation.

    The problem then is the suggestion in the question that the eigenvectors 'do not depend on the choice of basis'. That suggestion is meaningless, because no basis is used to determine the n-tuple of numbers or the matrix. They are basis-independent representations. If we pre and post-multiply the matrix by matrices C-1 and C respectively, and pre-multiply a vector by C we have not changed the basis in which they are represented. We have changed the matrix and the vector.

    Hence the question was incorrect to try to connect the question to bases. It would have been correct to instead ask 'show that, for any matrix ##A## and invertible matrix ##C##, the matrix ##C^{-1}AC## has the same eigenvalues as ##A## but not necessarily the same eigenvectors.'*

    By mentioning bases, the question conflates the notion of an n-tuple of numbers as a vector in vector space ##\mathbb{R}^n## with the notion of an n-tuple as a representation of a vector in ##\mathbb{R}^n## (or in any other n-dimensional vector space over the reals), and the two are entirely different things.

    Such confusion would prove particularly unhelpful when one starts to do Hamiltonian mechanics and wishes to distinguish carefully between active and passive transformations.

    * As well as removing the confusion, that also points towards an easy solution. One simply has to show that if the pair ##(\lambda,\vec{v})## is a solution of
    $$C^{-1}AC\vec{v}=\lambda\vec{v}$$
    then there exists a vector ##\vec{u}## such that ##(\lambda,\vec{u})## is a solution of
    $$A\vec{u}=\lambda\vec{u}$$
    As Fredrik points out, there is no need to use determinants. The proof is two easy steps and of course one finds that ##\vec{u}=C\vec{v}##.
     
    Last edited: Sep 21, 2015
  15. Sep 22, 2015 #14
  16. Sep 22, 2015 #15

    Fredrik

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    Yes, that.
     
  17. Sep 22, 2015 #16
    Cheers!!!
     
  18. Sep 22, 2015 #17
    How do I show eigenvectors are not invariant and why the dependence of the eigenvectors in the particular basis is exactly what we expected and argue that in some sense they are indeed invariant???
     
  19. Sep 22, 2015 #18

    Fredrik

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    It sounds like you're asking us to rewrite the solutions that you've found (which are essentially complete already) so that they're 100% ready to hand in to your teacher. We provide hints, not complete solutions. You have to show us your work if you want more help.
     
  20. Sep 22, 2015 #19

    andrewkirk

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    deleted
     
  21. Sep 22, 2015 #20

    mfb

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    To show that something is not invariant you can give an example.
     
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