Results 1 to 15 of 15

Math Help - P-1 A P formula question

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    P-1 A P formula question

    A , B are 2x2 matrices in which
    det A=det B=tr(A)=tr(B)=7
    tr- is trace ,the sum of the diagonal

    could we link A and B
    by
    B=P-1AP
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    re: P-1 A P formula question

    Quote Originally Posted by transgalactic View Post
    A , B are 2x2 matrices in which det A=det B=tr(A)=tr(B)=7 tr- is trace ,the sum of the diagonal could we link A and B by B=P-1AP
    Hint: The characteristic polynomials of A and B are \chi(\lambda)=\lambda^2-(\textrm{tr}A)\lambda+\det A=\lambda^2-7\lambda+7 .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: P-1 A P formula question

    how you got this polinomial
    in ordrer to get The characteristic polynomials of A is |A-KI|
    i dont have the matrices of A
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2010
    Posts
    193

    re: P-1 A P formula question

    Quote Originally Posted by transgalactic View Post
    how you got this polinomial
    in ordrer to get The characteristic polynomials of A is |A-KI|
    i dont have the matrices of A
    He showed you how he got the polynomial. The trace and determinant of a matrix show up as particular coefficients of the characteristic polynomial of that matrix.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,324
    Thanks
    699

    re: P-1 A P formula question

    if we have a 2x2 matrix: A =

    [a b]
    [c d]

    then det(A - xI) = (a-x)(d-x) - bc = ad - (a+d)x + x^2 - bc = (ad - bc) - (a+d)x + x^2 = det(A) - tr(A)x + x^2.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: P-1 A P formula question

    ok i have two eigen values in 2x2 matric
    thus making A diagonisable

    but it doesnt say that the digonolised form of A is B

    ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2010
    Posts
    193

    re: P-1 A P formula question

    Quote Originally Posted by transgalactic View Post
    ok i have two eigen values in 2x2 matric
    thus making A diagonisable

    but it doesnt say that the digonolised form of A is B

    ?
    Well, by the same exact reasoning, wouldn't B also be diagonalizable? And wouldn't A and B have the same diagonalized forms, since they have the same eigenvalues?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2

    re: P-1 A P formula question

    Quote Originally Posted by topspin1617 View Post
    Well, by the same exact reasoning, wouldn't B also be diagonalizable? And wouldn't A and B have the same diagonalized forms, since they have the same eigenvalues?

    In general no, by no means. BUT in the very particular case of 2x2 matrices yes.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2010
    Posts
    193

    re: P-1 A P formula question

    Quote Originally Posted by tonio View Post
    In general no, by no means. BUT in the very particular case of 2x2 matrices yes.

    Tonio
    Wait... what are we saying "no" to in general?

    I just meant that here, the matrices both have the same set of 2 distinct eigenvalues, so that both will be similar to \mathrm{diag}(\lambda_1,\lambda_2}).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: P-1 A P formula question

    i wasnt asked to compare the digonolised forms.

    i was asked to show that there is P in which B=P^-1 A P
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    re: P-1 A P formula question

    Similarity of matrices is an equivalence relation so, A\sim D and B\sim D implies A\sim B i.e. P^{-1}AP=B for some P .
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Nov 2010
    Posts
    193

    re: P-1 A P formula question

    Yeah... it's kind of a basic fact that can be proved with arithmetic...

    P^{-1}AP=Q^{-1}BQ\Rightarrow QP^{-1}APQ^{-1}=B\Rightarrow (PQ^{-1})^{-1}A(PQ^{-1})=B.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2

    re: P-1 A P formula question

    Quote Originally Posted by topspin1617 View Post
    Wait... what are we saying "no" to in general?

    I just meant that here, the matrices both have the same set of 2 distinct eigenvalues, so that both will be similar to \mathrm{diag}(\lambda_1,\lambda_2}).

    This is not what I understood: you said "And wouldn't and have the same diagonalized forms, since they have the same eigenvalues?", and

    to this ia answered "no, in the general case, but yes in the case of dimension 2"...that's all.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: P-1 A P formula question

    Quote Originally Posted by topspin1617 View Post
    Yeah... it's kind of a basic fact that can be proved with arithmetic...

    P^{-1}AP=Q^{-1}BQ\Rightarrow QP^{-1}APQ^{-1}=B\Rightarrow (PQ^{-1})^{-1}A(PQ^{-1})=B.
    PQ^-1 have to be invertible
    does multiplication of two invertables is invertible too?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,324
    Thanks
    699

    re: P-1 A P formula question

    yes, if A is invertible, and B is invertible, then AB is invertible.

    why? since we can form the matrices A^{-1} and B^{-1}, the product B^{-1}A^{-1} exists.

    but (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1}  =AIA^{-1} = AA^{-1} = I,

    and (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question: Quadratic Formula
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 19th 2010, 01:59 PM
  2. Formula Question
    Posted in the Algebra Forum
    Replies: 13
    Last Post: March 12th 2009, 08:05 AM
  3. Formula Question.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 11th 2009, 08:29 PM
  4. formula question
    Posted in the Algebra Forum
    Replies: 12
    Last Post: April 28th 2008, 04:04 AM
  5. [SOLVED] Formula question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: July 10th 2005, 10:23 PM

Search Tags


/mathhelpforum @mathhelpforum