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Find all Complex 2x2 matrices A

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Find all Complex 2x2 matrices A such that A2 = A

    2. Relevant equations
    [itex]
    \[ \left( \begin{array}{cc}
    a & b \\
    c & d \\ \end{array} \right)^2\] = A

    [/itex]
    3. The attempt at a solution

    A2=A

    I need to solve the equations

    [itex]
    a^2+bc=a
    [/itex]
    [itex]

    (a+d)b =b
    [/itex]
    [itex]

    (a+d)c = c\\
    [/itex]
    [itex]

    d^2 + bc = d\\
    [/itex]
    What is the best way to do this?

    Should I use some kind of rearrangement like

    using equation 1 and 4 we get

    [itex]
    a^2-d^2=a-d [/itex]

    any help greatly appreciated
     
  2. jcsd
  3. Oct 22, 2009 #2

    LCKurtz

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    I would think about two cases, that where A-1 exists and where it doesn't exist.
     
  4. Oct 23, 2009 #3
    Are you talking about when A is an orthogonal matrix so A-1 = AT?
     
    Last edited: Oct 23, 2009
  5. Oct 23, 2009 #4

    Office_Shredder

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    The question is: Is A invertible? If so, then just take A-1 of both sides and what do you get?
     
  6. Oct 23, 2009 #5
    A-1 exist when [itex] ad-cb \neq 0 [/itex]

    so
    case 1 [itex] ad-cb \neq 0 [/itex]
    case 2 [itex] ad-cb = 0 [/itex]
     
  7. Oct 23, 2009 #6

    Office_Shredder

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    Stop looking at the series of equations you have

    If
    A2=A

    And A-1 exists

    In one step you can find A
     
  8. Oct 23, 2009 #7
    If A is invertable and A2=A
    AA=A
    A-1AA=A-1A
    IA=I

    So A = I
     
  9. Oct 23, 2009 #8

    Office_Shredder

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    Now if A is not invertible, what can you say about how A acts on im(A), its image?
     
  10. Oct 23, 2009 #9

    HallsofIvy

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    Think about the fact that if [itex]A^2= A[/itex] then [itex]A^2- A= A(A- I)= 0[/itex]
     
  11. Oct 25, 2009 #10
    Would that mean that the only way that
    [itex]A^2- A= A(A- I)= 0[/itex]
    is that either A = the trivial solution.

    [itex]\[ \left( \begin{array}{cc}0 & 0 \\0 & 0 \\ \end{array} \right)\] = A[/itex]
    or

    [itex]\[ \left( \begin{array}{cc}-1 & 0 \\0 & -1 \\ \end{array} \right)\] = A[/itex]

    regards
     
    Last edited: Oct 25, 2009
  12. Oct 25, 2009 #11

    Dick

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    That's completely wrong thinking. If A and B are matrices AB=0 does NOT mean A=0 or B=0. And [[-1,0],[0,-1]] doesn't work at all. Here's some other ones that do [[1,0],[0,0]], [[1,1],[0,0]], [[1/2,1/2],[1/2,1/2]]. There's LOTS of them. You already followed up on the excellent hint that if A is invertible, then A=I is the only solution. Now suppose A is not invertible. Then det(A)=0, so you can add the condition ad-bc=0 to your original list of equations. That makes it easier. Now go to those equations and start working on cases. You should be able to write down a two parameter family of solutions that satisfy A^2=A besides the zero matrix.
     
  13. Oct 25, 2009 #12

    Hurkyl

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    For the record, responders have been suggesting two different methods of approach. Office_Shredder's is a geometric one, and offers more insight into why idempotent matrices are interesting. The other was more algebraic.


    I confess I would have suggested yet a different approach, one that starts by looking at the equation (a+d)b=b.
     
  14. Oct 25, 2009 #13

    Dick

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    Why is that different from the algebraic approach? I'm just saying that the equations the OP originally posted are a lot easier to solve if you restrict to the case of noninvertible matrices, in which case you can put ad=bc. LCKurtz already suggested this, and it's a very good suggestion.
     
    Last edited: Oct 25, 2009
  15. Oct 25, 2009 #14
    As idempotent matrices must eigenvalues of 1 and 0 would it be all the Matrices that can produce the diagonal matrix

    [itex]\[ \left( \begin{array}{cc}1 & 0 \\0 & \\ \end{array} \right)\] = A[/itex]
     
  16. Oct 25, 2009 #15

    Dick

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    No, I already pointed out that [[1/2,1/2],[1/2,1/2]] works.
     
  17. Oct 25, 2009 #16

    Hurkyl

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    Figures you'd go for yet another solution! There are (at least) three things you need to do to turn this into a complete solution and a proof.

    (1) What about non-diagonalizable matrices?

    (2) For the diagonalizable solutions, you haven't covered all possible diagonal forms.

    (3) You still need to simplify your characterization of the solutions into a form that would be considered "found".
     
  18. Oct 30, 2009 #17
    So I have the first case where the matreix is invertable


    If A is invertable and A^2=A
    AA=A
    A-1AA=A-1A
    IA=I

    So A = I

    The other case is when A is not invertable so then the det(A) = 0

    A^2 = A
    AA = A
    AA - A = A - A
    AA - A = 0
    A(I-A) = 0


    SO what complex matrix will do this?
     
  19. Oct 30, 2009 #18

    Dick

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    I already told you in post 10 that that factorization isn't particularly useful. Go back to your original set of equation and add the condition det(A)=0=ad-bc. You'll find they are pretty easy to solve.
     
  20. Nov 4, 2009 #19
    For det(A)= 1 = invertable

    [itex]\[ \left( \begin{array}{cc}1 & 0 \\0 & 1 \\ \end{array} \right)\] [/itex]


    for det(A) = 0 = non invertable

    b or c can be any value

    [itex]\[ \left( \begin{array}{cc}1 & b \\0 & 0 \\ \end{array} \right)\] [/itex]

    [itex]\[ \left( \begin{array}{cc}1 & 0 \\c & 0 \\ \end{array} \right)\] [/itex]

    [itex]\[ \left( \begin{array}{cc}0 & b \\0 & 1 \\ \end{array} \right)\] [/itex]

    [itex]\[ \left( \begin{array}{cc}0 & 0 \\c & 0 \\ \end{array} \right)\] [/itex]

    [itex]\[ \left( \begin{array}{cc}0 & 0 \\0 & 0 \\ \end{array} \right)\] [/itex]

    also you have
    for all square matrices nXn

    [itex]\[ \left( \begin{array}{cc}\frac{1}{n} & \frac{1}{n} \\\frac{1}{n} & \frac{1}{n} \\ \end{array} \right)\] [/itex]


    I'm not sure how to get that last one. I just noticed that DIck posted that one and by trying different options noticed that it worked for all nXn matrices
     
  21. Nov 4, 2009 #20

    Dick

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    Those are all one parameter special cases. At least the ones that actually work are. The really general solution has two free parameters b and c, both in the same matrix. Can you find it? It includes the [[1/2,1/2],[1/2,1/2]] case. I keep telling you to go back to your original equations and add ad=bc.
     
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