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A simple yet complicated matrix problem

  1. Jan 7, 2009 #1
    The problem is:

    Let K = a 2x2matrix where a is a real number and does not equal zero. That is, all 2x2
    in which each entry is the same. Show that is a group under multiplication. And find its identity element. Verify. Please note: That the basic idenity matrix is not the identity element

    I have been looking through even book I own, but if all the entries are the same, couldn't that make the matrix singular?
     
  2. jcsd
  3. Jan 7, 2009 #2

    Dick

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    Sure, the matrices are all singular. But that doesn't mean they can't form a group. The identity of that group is also singular. Can you find it? If E is the identity you want to show that AE=EA=A for all matrices with constant entries.
     
  4. Jan 7, 2009 #3

    HallsofIvy

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    Oh, thats a cute problem! Certainly, my first reaction was "Why, that can't be true!" So it's a good thing Dick answered first.

    crazoyformath2, the point is that the identity matrix and inverse matrices are not the same for this set as they would be for all matrices.

    [tex]\begin{bmatrix}a & a \\ a & a\end{bmatrix}[/tex]
    and
    [tex]\begin{bmatrix}b & b \\ b & b\end{bmatrix}[/tex]
    are in that set. What is the product of those two matrices? What must "a" be in order that that product be exactly
    [tex]\begin{bmatrix}b & b \\ b & b\end{bmatrix}[/tex]?
     
  5. Jan 7, 2009 #4
    I am so lost. I understand what Dick said to a point. I know the determinant cannot be 0 for then there would not be an inverse, correct? but if all the entries are the same won't the determinate be zero?
     
  6. Jan 7, 2009 #5

    Dick

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    This isn't about determinant and the usual formula for the inverse. The 'inverse' in your group is not the same as the inverse in the usual sense. Because the 'identity' is not the usual identity. If your group is going to have an identity, it has to have all matrix entries equal. So it can't be [[1,0],[0,1]]. There is another matrix that can act as the identity in this group. Stop fretting about this so much and do the exercise Hall's proposes.
     
  7. Jan 7, 2009 #6
    Remember what identity means, it's an element e such that e*A = A*e = A (where * is the operation in your group) for all A in your group. So what Halls proposed makes perfect sense, pick an arbitrary element of your group , namely the 2x2 matrix with a's in every entry where a is a real number (which is what elements of this group look like). Then we are trying to find the identity element which will also look like a 2x2 matrix with b's in every entry, now multiply them together, and you should get back your 2x2 matrix with a's in every entry. So at this point it's just matrix multiplication.
     
  8. Jan 7, 2009 #7
    wouldnt "a" have to be all one's?
     
  9. Jan 7, 2009 #8
  10. Jan 7, 2009 #9

    Dick

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    No. Did you actually try multiplying a matrix that is all ones times another matrix with constant entries? Why isn't it an identity?
     
  11. Jan 7, 2009 #10
    nevermind i really am an idiot with these matrices, when i multipy a matrices of all a's but a matrice of all b's, i get ab+ab in each entry. and when i do the opposite, i'll get the same answer, but i just don't understand how i'd get back to a.
     
  12. Jan 7, 2009 #11

    Dick

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    Sure. You get 2ab in each entry. How can 2ab=a?
     
  13. Jan 7, 2009 #12
    if b = 1/2
     
  14. Jan 7, 2009 #13

    Dick

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    You've got it. So the 'identity' matrix in your group is [[1/2,1/2],[1/2,1/2]]. Can you figure out how to find the 'inverse' of a matrix [[a,a],[a,a]]?
     
  15. Jan 7, 2009 #14
    i know i don't work it out like a normal matrix right?
     
  16. Jan 7, 2009 #15

    Dick

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    If you mean determinants and stuff, no. Let's call the matrix with entries all a, M(a). So far you figured out that M(1/2) is the identity. So M(1/2)*M(a)=M(a). If M(b) is the inverse of M(a) then you need to solve M(a)*M(b)=M(1/2), right? Just solve for b in terms of a, like you did before.
     
  17. Jan 7, 2009 #16
    so M(b) would be a matrix of all a/4 right?? and then M(a) * M(b) = M(1/2)
     
  18. Jan 7, 2009 #17

    Dick

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    I get M(a)*M(a/4)=M(a^2/2). That's not quite it, right?
     
  19. Jan 7, 2009 #18
    Aren't you basically trying to solve
    [tex]\begin{bmatrix}a & a \\ a & a\end{bmatrix} \begin{bmatrix}c & c \\ c & c\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{bmatrix}[/tex]
     
  20. Jan 7, 2009 #19
    yeaaa...!!
     
  21. Jan 7, 2009 #20

    Dick

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    Ok, so you got it, right? M(a)*M(1/(4a))=M(1/2)?
     
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