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Linear Algebra: Idemponent matrix

  1. Sep 14, 2009 #1
    In idemponent matrices:
    Is there a value of r for which
    (matrix) A= -1 -1
    2 r

    is idempotent? If so, what is the value of r and why.

    I know that A^2=A, that's about all... How does one find r, if this is the only info given?

    I have no idea how to approach this. Please help.
     
    Last edited: Sep 14, 2009
  2. jcsd
  3. Sep 14, 2009 #2
    What is an indemponent matrix?

    Thanks
    Matt
     
  4. Sep 14, 2009 #3
    If A is an n x n matrix, then A is called idempotent if A^2=A
     
  5. Sep 14, 2009 #4
    Sorry, can't help you with that. But thanks for letting me no what idempotent means.

    Thanks
    Matt
     
  6. Sep 14, 2009 #5

    gabbagabbahey

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    The simple definition is simply a matrix for which [itex]A^2=A[/itex].


    Well, if [tex]A=\begin{pmatrix}-1 & 1 \\ 2 &r \end{pmatrix}[/tex], what will [itex]A^2[/itex] be?....Setting [itex]A^2=A[/itex] should give you 4 equations (one for each component of the matrices), is there a value of [itex]r[/itex] that solves all 4 equations simultaneously?
     
  7. Sep 14, 2009 #6
    A^2 = -1 -1
    2 r

    and -1 -1 = -1 -1
    2 r 2 r ???
     
  8. Sep 14, 2009 #7
    sorry for bad layout... still haven't figured out how to use advanced reply
     
  9. Sep 14, 2009 #8
    Think about the determinant of an idempotent matrix.
     
  10. Sep 14, 2009 #9

    gabbagabbahey

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    No,

    [tex]A=\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}\implies A^2=\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}[/tex]

    Carry out the matrix multiplication.
     
  11. Sep 14, 2009 #10
    ok I get:
    -1 1-r
    -2+2r -2+r2
     
  12. Sep 14, 2009 #11

    gabbagabbahey

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    Right, and for [itex]A[/itex] to be idempotent, that must also be equal to the matrix [itex]A[/itex]...so you want to find an [itex]r[/itex] such that

    [tex]\begin{pmatrix}-1 & 1-r \\ -2+2r &-2+r^2 \end{pmatrix}=\begin{pmatrix}-1 & -1 \\ 2 &r \end{pmatrix}[/tex]
     
  13. Sep 14, 2009 #12
    ok thank you... do I now have to reduce it to row-echelon form, by using the Gaussian Elimination?
     
  14. Sep 14, 2009 #13

    HallsofIvy

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    You don't have to. If [itex]A^2= A[/itex] then you must have -1= -1, -1= 1-r, -2+2r= 2 and -2+r^2= r. Are there values of r that satisfy all of those and, if so, what are they?
     
  15. Sep 14, 2009 #14
    r= 2; r=1/2; r=-1
     
  16. Sep 14, 2009 #15

    gabbagabbahey

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    r=2 works, but the other two values do not satisfy all 4 equations. (For the two matrices to be equal, all 4 of there components must be equal)
     
  17. Sep 15, 2009 #16
    thank you so much for your help. One last question on the topic: is a square invertible idempotent matrix also the Identity matrix?
     
  18. Sep 15, 2009 #17

    gabbagabbahey

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    Well, if [itex]A^2=A[/itex], what can you say about [itex]\detA[/itex]?
     
  19. Sep 15, 2009 #18
    I don't understand the last part of what you said
     
  20. Sep 15, 2009 #19

    gabbagabbahey

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    Take the determinant of both sides od the equation [itex]A^2=A[/itex]....what do you get?
     
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