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Matrices and eigen value problems

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Let A and B be nxn matrices over reals. Show that I - BA is invertible if I - AB is invertible. Deduce that AB and BA have the same eigen values

    2. Relevant equations

    det(AB) = det(A).det(B)

    3. The attempt at a solution

    given: (I-AB) is invertible

    -> det(I-AB) is not equal to 0

    i.e. (-1)^n times det(AB-I) is not equal to 0
    i.e. (-1)^n times det(AB)-det(I) is not equal to 0
    i.e. (-1)^n times det(A).det(B)-det(I) is not equal to 0
    i.e. (-1)^n times det(B).det(A)-det(I) is not equal to 0
    i.e. (-1)^n times det(BA)-det(I) is not equal to 0
    i.e. (-1)^n times det(BA-I) is not equal to 0

    -> det(I-BA) is not equal to 0

    and hence (I - BA) is invertible, if (I - AB) is invertible

    as regards to the second part i have not got it yet...
  2. jcsd
  3. Aug 9, 2011 #2
    While the determinant may be linear in the rows and columns of a matrix, it is in general NOT linear with respect to entire matrices. That is, [itex] \det (A+B) \neq \det A + \det B[/itex].

    Hint: take a look at the matrix [itex]B(I-AB)^{-1}A+I[/itex].
  4. Aug 9, 2011 #3
    Also, for the eigenvalues, try multiplying the eigenvalue/vector equation [itex]ABx = \lambda x[/itex] by a certain matrix to transform it into an equation involving BA.
  5. Aug 9, 2011 #4
    i am just a little confused here.. see..

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  6. Aug 10, 2011 #5
    Make up two 2x2 matrices. Odds are they they will provide a counterexample to

    [tex] \det (A+B) = \det A +\det B.[/tex]

    Or, for a more general argument with 2x2's, suppose

    [tex] A= \begin{bmatrix} \leftarrow & a_1 &\rightarrow \\ \leftarrow & a_2 & \rightarrow\end{bmatrix}, \quad B= \begin{bmatrix} \leftarrow & b_1 &\rightarrow \\ \leftarrow & b_2 & \rightarrow\end{bmatrix}[/tex]

    where [itex]a_1,a_2[/itex] and [itex] b_1,b_2[/itex] are the rows of A and B, respectively. Then

    [tex] \det (A+B) = \det \begin{bmatrix} a_1+b_1 \\ a_2+b_2\end{bmatrix} = \det \begin{bmatrix}a_1 \\a_2+b_2\end{bmatrix}+\det\begin{bmatrix}b_1 \\a_2+b_2\end{bmatrix}=\det\begin{bmatrix}a_1 \\a_2\end{bmatrix}+\det\begin{bmatrix}a_1 \\b_2\end{bmatrix}+\det\begin{bmatrix}b_1 \\a_2\end{bmatrix}+\det\begin{bmatrix}b_1 \\b_2\end{bmatrix} = \det A + \det B +\det\begin{bmatrix}a_1 \\b_2\end{bmatrix}+\det\begin{bmatrix}b_1 \\a_2\end{bmatrix}. [/tex]

    If those extra terms vanish, then [itex] \det (A+B) =\det A +\det B[/itex]. But in general, it's not true. It would be nice if it were though. :smile: Remember it's ONLY linear in the rows and columns, not the entire matrix itself. This is also the reason that if A is nxn, then

    [tex] \det (cA) = c^n \det A.[/tex]
  7. Aug 10, 2011 #6
    thanks a lot.. any idea then how to solve the first part of the problem.. i am stuck...
  8. Aug 10, 2011 #7
    Did you take a look at the hint in my first post? Try verifying that that matrix is the inverse you're looking for.
  9. Aug 10, 2011 #8
    ya .. well i didnt understand .. i wish you could break it down for me.. and how you got the idea to solve it... is there any particular book thats good for "matrices" and "linear algebra" . . . ?
  10. Aug 10, 2011 #9
    also could you help me out with this problem too..

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  11. Aug 11, 2011 #10
    For future reference, start a new thread when you have a new question. You'll get a lot more traffic and lots more help. :smile:

    EDIT: I wrote a big long post, but then forgot to ask the most important question first! Can you show me what you've tried so far?
    Last edited: Aug 11, 2011
  12. Aug 11, 2011 #11
    well what i have tried so far has been the wrong approach as you have pointed out.. thanks for that ..... so to be honest ... right now i have nothing to show .. as i didn't understand how really B(I−AB)^−1.A+I is the inverse to the matrix i am looking for??????
  13. Aug 11, 2011 #12
    There are a variety of ways to show that a matrix is invertible: show it's determinant is different from zero, show that it has full rank, show that the equation Ax=0 has only the trivial solution, etc. etc. However, we're going to construct a matrix that will send I-BA to the identity, and since inverses are unique, this must be the inverse we're after. Also note that--in the case of matrices at least--showing a matix is a left- or right-inverse is equivalent to showing that it is THE two-sided inverse.

    So we need to show that

    [tex] (I-BA)(B(I-AB)^{-1}A+I) = I.[/tex]

    That matrix, our candidate for the inverse, exists by hypothesis, i.e., it contains the inverse of I-AB which we are allowed to assume exists. So, therefore, the first step is to expand that multiplication:

    [tex](I-BA)(B(I-AB)^{-1}A+I) = B(I-AB)^{-1}A+I -BA(B(I-AB)^{-1}A+I) = B(I-AB)^{-1}A+I -BAB(I-AB)^{-1}A-BA.[/tex]

    Try taking over from here...
  14. Aug 12, 2011 #13
    very interesting thanks .. a lot .. could you tell me of any good book which covers matrices thoroughly ???? thanks again . .
  15. Aug 12, 2011 #14
    It's hard to know what exactly to recommend when I don't know your background.

    There is a subforum over in Academic Guidance called Science Book Discussion where they will recommend books. Maybe you should post over there. Or, I'm sure there will probably be some results that pop up if you use the search feature. And, take a look https://www.physicsforums.com/blog.php?b=3206" [Broken].

    I didn't learn my linear algebra and matrix theory all from one source, it was kind of a piecemeal education that I picked up as I went along. I hated my intro linear algebra text and course, so perhaps that's why. Now it's one of my favorite things to talk about.
    Last edited by a moderator: May 5, 2017
  16. Aug 14, 2011 #15
    sorry for the late reply... thanks for all the help ...:smile:
    Last edited by a moderator: May 5, 2017
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