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Linear Algebra-Inverses

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine which of the formulas hold for all invertible nxn matrices A and B

    1. (In-A)(In+A) = In - A^2
    2. (A+B)(A-B) = A^2 - B^2
    3. A^8*B^5 is invertible
    4. (A+A^-1)^9 = A^9 + A^-9
    5. AB = BA
    6. A+A^-1 is invertible

    I know 5 is right, and number 2 wrong, but what else?
     
  2. jcsd
  3. Feb 27, 2007 #2

    radou

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    A starting hint: a product of regular matrices is a regular matrix.
     
  4. Feb 27, 2007 #3
    I know that but that doesn't solve the problem
     
  5. Feb 28, 2007 #4

    HallsofIvy

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    For 1, 4, and 6, just go ahead and do the multiplication! What is (I-A)(I+A)?
    If you don't want to actually multiply (A+A-1) nine times, try one or two times and see if you can't find a pattern. That's how you solve math problems- you try things, you don't just sit and try to "remember" how to solve it! As for (A+ A-1) try to guess an inverse and then do the calculation to see what happens. Try it with a simple diagonal matrix first. (Another general math method- to do complex problems, try a few simple examples first.)
     
  6. Feb 28, 2007 #5
    I sub n is an identity matrix. 2, 3 and 5 seems to work when using a diagonal matrix.
     
  7. Feb 28, 2007 #6

    radou

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    You're not interested in diagonal matrices, you're interested in all regular (i.e. invertible) matrices. Take, for example [tex]A=\left(\begin{array}{cc}1 & 2\\0 & -1\end{array}\right)[/tex], and [tex]B=\left(\begin{array}{cc}-1 & 0\\2 & 3\end{array}\right)[/tex]. Obviously [tex]AB\neq BA[/tex], although both det(A) and det(B) are non-zero.

    Edit: after reading this, you should easily see if (2) holds or not for two regular nxn matrices A and B.
     
    Last edited: Feb 28, 2007
  8. Feb 28, 2007 #7
    Thanks radou, after some thought, I figured out that 1 and 3 are the only formulas that hold for all invertible n x n matrices A and B.
     
  9. Mar 1, 2007 #8

    radou

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    Assuming that by " (In-A)(In+A) = In - A^2 " you meant " (In-A)(In+A) = In^2 - A^2 ", where In = A^-1.
     
  10. Mar 1, 2007 #9

    HallsofIvy

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    In= A-1? Why would you assume that? Craigy specifically said In is the n by n identity matrix.
    (In- A)(In+ A)= In2+ InA- AIn+ A2. Since InA= AIn= A, and In2= In, yes, (In- A)(In+ A)= In- A2.
     
  11. Mar 1, 2007 #10

    radou

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    The notation confused me, it's okay now.
     
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