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  1. #1
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    how do i..

    How do I find a matrix P, where there is a matrix A, such that P^(-1)AP is diagonal??
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  2. #2
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    Quote Originally Posted by kgpretty View Post
    How do I find a matrix P, where there is a matrix A, such that P^(-1)AP is diagonal??
    Look up the eigen decompositon either here or here

    RonL
    Last edited by CaptainBlack; October 8th 2006 at 10:30 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Look up the eigen decompositon either here or here

    RonL
    Thankyou very much. I like the explanation wikipedia gave better.
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  4. #4
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    Another question asks, for a matrix A, show that it satisfies it's characteristic equation.

    Looking at the example on Wikipedia, would the characteristic equation be (Y - 3)(Y - 2)(Y - 1)?

    If it is, how do I complete the above question?
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  5. #5
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    Quote Originally Posted by kgpretty View Post
    Another question asks, for a matrix A, show that it satisfies it's characteristic equation.

    Looking at the example on Wikipedia, would the characteristic equation be (Y - 3)(Y - 2)(Y - 1)?

    If it is, how do I complete the above question?
    The charateristic equation is:

    det(A-lambda I) = 0

    (as a polynomial in lambda). Now as 3, 2, 1 are the roots of this equation
    we know that if we expand the above expression we will have:

    (lambda-3)(lambda-2)(lambda-1)=0

    for the charateristic equation, and as A should satisfy this we should find
    that:

    (A-3)(A-2)(A-1)=0

    where there is an impicit multiplication of the 3, 2, 1 and 0 by I.

    This expands to:

    (A-3)(A-2)(A-1) = (A-3)(A^2 - 3A +2) = A^3 - 6A^2 + 11A -6

    The powers of A are simply A^2=AA, A^3=A(AA), and the 6 is short hand
    for 6 I.

    RonKL
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  6. #6
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    Does the question stop at A^3 - 6A^2 + 11A -6
    Or do I have to break down the above equation even further. Like find A^3, 6A^2 and so on, then do the addition and substraction??
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  7. #7
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    Quote Originally Posted by kgpretty View Post
    Does the question stop at A^3 - 6A^2 + 11A -6
    Or do I have to break down the above equation even further. Like find A^3, 6A^2 and so on, then do the addition and substraction??
    I'm a bit confused by what you need to show. With a particular matrix you
    can compute the powers and substitute them into the equation and show
    that the matrix satisfies it.

    For a general matrix you could show this is in general true.

    If D is the diagonal matrix then as each element on the diagonal is an eigen
    value of A, we have tivialy that D satisfies the charateristic equation.

    Also as P D P^-1 = A, and P D^2 P^-1 = P D P^-1 P D P^-1 = A^2,
    and so on for higher powers, so P D^n P^-1 = A^n and we have:

    if

    sum_{r=0,n} a_r x^r = 0

    is a polynomial satisfied by D, then it is also satisfied by A, since:

    P sum_{r=0,n} a_r D^r P^-1 = sum_{r=0,n} a_r P D^r P^-1

    ..................=sum_{r=0,n} a_r A^r = 0.

    RonL
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  8. #8
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    Ok, I think I get it. Thankyou.

    However I have a question for which I am completely stumped.
    First, it asked you to find the eigenvectors and eigenvalues for matrix A.
    Code:
    Matrix A:
    |1 3|
    |2 6|
    
    Here's what I did:
    |1 3|
    |2 6|x  = Yx
    
    |1 3|
    |2 6|x - Yx = 0
    
    |1-Y      3  |
    |  2    6 - Y|x    = 0
    
    .:
    (1 - Y)(6 - Y) - 6 = 0
    6 + Y^2 - 6Y - 1Y - 6 = 0
    Y^2 - 7Y + 0 = 0
    (Y + 0)(Y - 7) = 0
    
    When Y = 0
    |1 3| |x1|
    |2 6| |x2| = 0
    
    Since
    |1 3| | 3|
    |2 6| |-1| = 0
    
    | 3|
    |-1| is an eigenvector corresponding to the eigenvalue 0.
    
    
    When Y = 7
    |1-7     3|
    |2     6-7|x = 0
    
    |-6   3| |x1|
    | 2  -1| |x2| = 0
    
    Since
    |-6   3| |1|
    | 2  -1| |2| = 0
    
    |1|
    |2| is an eigenvector corresponding to the 
    eigenvalue 7.
    
    
    Good, so I got the eigenvalues and vectors. 
    The question then asked to find P for which 
    P^-1AP is a diagonal matrix.
    P = |3   1|
        |-1  2|
    
    P-1 = |3   1|
          |-1  2|
    
    = | 3   1|1  0|
      |-1   2|0  1|
    
    = |1  0|2  -1|
      |0  1|1   3|
    
    = |2  -1|
      |1   3|
    
    .:
    P^-1AP = |2  -1|  |1  3|  | 3  1|
             |1   3|  |2  6|  |-1  2|
    
               = |0   0| |3   1|
                 |7  21| |-1  2|
    So you see my problem, P^-1AP is not equal to a diagonal.... All the other problems I did the same way gave me a diagonal.
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  9. #9
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    Quote Originally Posted by kgpretty View Post
    Ok, I think I get it. Thankyou.

    However I have a question for which I am completely stumped.
    First, it asked you to find the eigenvectors and eigenvalues for matrix A.
    Code:
    Matrix A:
    |1 3|
    |2 6|
     
    Here's what I did:
    |1 3|
    |2 6|x  = Yx
     
    |1 3|
    |2 6|x - Yx = 0
     
    |1-Y      3  |
    |  2    6 - Y|x    = 0
     
    .:
    (1 - Y)(6 - Y) - 6 = 0
    6 + Y^2 - 6Y - 1Y - 6 = 0
    Y^2 - 7Y + 0 = 0
    (Y + 0)(Y - 7) = 0
     
    When Y = 0
    |1 3| |x1|
    |2 6| |x2| = 0
     
    Since
    |1 3| | 3|
    |2 6| |-1| = 0
     
    | 3|
    |-1| is an eigenvector corresponding to the eigenvalue 0.
     
     
    When Y = 7
    |1-7     3|
    |2     6-7|x = 0
     
    |-6   3| |x1|
    | 2  -1| |x2| = 0
     
    Since
    |-6   3| |1|
    | 2  -1| |2| = 0
     
    |1|
    |2| is an eigenvector corresponding to the 
    eigenvalue 7.
     
     
    Good, so I got the eigenvalues and vectors. 
    The question then asked to find P for which 
    P^-1AP is a diagonal matrix.
    P = |3   1|
        |-1  2|
     
    P-1 = |3   1|
          |-1  2|
     
    = | 3   1|1  0|
      |-1   2|0  1|
     
    = |1  0|2  -1|
      |0  1|1   3|
     
    = |2  -1|
      |1   3|
     
    .:
    P^-1AP = |2  -1|  |1  3|  | 3  1|
             |1   3|  |2  6|  |-1  2|
     
               = |0   0| |3   1|
                 |7  21| |-1  2|
    So you see my problem, P^-1AP is not equal to a diagonal.... All the other problems I did the same way gave me a diagonal.
    If:

    P = [3 1]
    .....[-1 2]

    P^-1 =.[4/10 2/10]
    ...........[2/10 6/10]

    Ooops, I appear to have inverted the wrong matrix some how, the inverse
    should be:

    P^-1= [2/7 -1/7]
    .........[1/7...3/7]

    RonL
    Last edited by CaptainBlack; October 11th 2006 at 07:38 AM.
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  10. #10
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    Quote Originally Posted by CaptainBlack View Post
    If:

    P = [3 1]
    .....[-1 2]

    P^-1 =.[4/10 2/10]
    ...........[2/10 6/10]

    RonL
    How did you arrive at
    P^-1 = [4/10 2/10]
    [2/10 6/10]
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  11. #11
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    Quote Originally Posted by kgpretty View Post
    How did you arrive at
    P^-1 = [4/10 2/10]
    [2/10 6/10]
    some sort of brain fade has occured, I must have inverted the wrong matrix
    (this must be the case as I multiplied what I inverted by this "inverse" and
    got the identity matrix)

    Checking this, the real inverse is


    P^-1= [2/7 -1/7]
    .........[1/7...3/7]


    obtained by Gaussian elimination with multiple right hand sides.

    There are many ways of inverting a 2x2 matrix, there is Gaussian elimination,
    Cramers rule, inspection, ..

    Code:
     
    Demo of Gaussian elimination
    ----------------------------
     
    Initial set-up:
     
     3 1  |  1  0
    -1 2  |  0  1
     
    multiply the second row by 3 and add the first:
     
     3 1  |  1  0
     0 7  |  1  3
     
    multiply the first row by 7 and subtract the second:
     
    21 0  |  6 -3
     0 7  |  1  3
     
    Divide the first row through by 21, and the second by 7:
     
     1 0  |  2/7  -1/7
     0 1  |  1/7   3/7
     
    So the inverse the the matrix:
     
    [2/7  -1/7]
    [1/7   3/7]
     
    Check:
     
    >  p=[3,1;-1,2]
                3             1 
               -1             2 
    >ip=[2/7,-1/7;1/7,3/7]
         0.285714     -0.142857 
         0.142857      0.428571 
    >
    >
    >ip.p
                1             0 
                0             1 
    >
    RonL
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  12. #12
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    Quote Originally Posted by CaptainBlack View Post
    some sort of brain fade has occured, I must have inverted the wrong matrix
    (this must be the case as I multiplied what I inverted by this "inverse" and
    got the identity matrix)

    Checking this, the real inverse is


    P^-1= [2/7 -1/7]
    .........[1/7...3/7]


    obtained by Gaussian elimination with multiple right hand sides.

    There are many ways of inverting a 2x2 matrix, there is Gaussian elimination,
    Cramers rule, inspection, ..

    Code:
     
    Demo of Gaussian elimination
    ----------------------------
     
    Initial set-up:
     
     3 1  |  1  0
    -1 2  |  0  1
     
    multiply the second row by 3 and add the first:
     
     3 1  |  1  0
     0 7  |  1  3
     
    multiply the first row by 7 and subtract the second:
     
    21 0  |  6 -3
     0 7  |  1  3
     
    Divide the first row through by 21, and the second by 7:
     
     1 0  |  2/7  -1/7
     0 1  |  1/7   3/7
     
    So the inverse the the matrix:
     
    [2/7  -1/7]
    [1/7   3/7]
     
    Check:
     
    >  p=[3,1;-1,2]
                3             1 
               -1             2 
    >ip=[2/7,-1/7;1/7,3/7]
         0.285714     -0.142857 
         0.142857      0.428571 
    >
    >
    >ip.p
                1             0 
                0             1 
    >
    RonL

    I got the same matrix elements for p^-1 as you did.
    You have shown that ip.p is a diagonal matrix. However, the question asked to show P^-1AP is a diagonal and that's where I got stumped because it isn't.
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