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Math Help - Find the square root of a 3x3 matrix

  1. #1
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    Find the square root of a 3x3 matrix

    I apologize if my notation isn't clear, newbie to this forum

    I'm trying to find out how to find the square root of a 3x3 matrix.

    For A=

    [1, 1, 1
    0, 1, 1
    0 , 0, 1]

    I know that, in general, A^x = (P^-1) (D^x) (P) for some invertible P. In the case of linearly independent eigenvectors P should form a basis of A's eigenspace. But, the eigenvalues of A here are all 1, and only has one eigenvector, [1, 0, 0] and its scalar multiples. So that method isn't going to work.

    There is a method using Spectral Decomposition that I don't fully understand. It starts with the equation

    for A nxn, eigenvalues v1....vs, multiplicities m1....ms, then there exists n uniquely defined consituent matrices E i,k: i = 1...s, k = 0.... m-1
    s.t. for any analytic function f(x) we have

    f (A) = (s sigma i = 1) (mi sigma k =0) f^(k) (vi) E i,k

    Anyways if you can decode that it seems to me you can arrive at the constituent matrices of A by the following equations

    (A-I) (A-I) = 0 + 0 + 2 E 1,0
    (A-I) (A-I) = 2 E 1,2
    which works out to
    [ 0, 0, 1/2
    0, 0, 0
    0, 0, 0 ]

    A-I = E 1,1
    which is of course
    [ 0, 1, 1
    0, 0, 1
    0, 0, 0]

    and finally

    I = E1,0

    So we have 3 constituent matrices for A, let's say

    X E1,0 + Y E 1,1 + Z E 1,2

    It turns out for values X=1, Y= 1/2, and Z = -1/4 you get

    [ 1, 1/2, 3/8
    0, 1, 1/2
    0, 0, 1]

    whose square is A. So somehow (I don't know) we have to use the constituent matrices in a linear equation to general the square root of A. How to get the values of X,Y,Z I do not know.
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  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Gchan View Post
    I apologize if my notation isn't clear, newbie to this forum

    I'm trying to find out how to find the square root of a 3x3 matrix.

    For A=

    [1, 1, 1
    0, 1, 1
    0 , 0, 1]

    I know that, in general, A^x = (P^-1) (D^x) (P) for some invertible P. In the case of linearly independent eigenvectors P should form a basis of A's eigenspace. But, the eigenvalues of A here are all 1, and only has one eigenvector, [1, 0, 0] and its scalar multiples. So that method isn't going to work.

    There is a method using Spectral Decomposition that I don't fully understand. It starts with the equation

    for A nxn, eigenvalues v1....vs, multiplicities m1....ms, then there exists n uniquely defined consituent matrices E i,k: i = 1...s, k = 0.... m-1
    s.t. for any analytic function f(x) we have

    f (A) = (s sigma i = 1) (mi sigma k =0) f^(k) (vi) E i,k

    Anyways if you can decode that it seems to me you can arrive at the constituent matrices of A by the following equations

    (A-I) (A-I) = 0 + 0 + 2 E 1,0
    (A-I) (A-I) = 2 E 1,2
    which works out to
    [ 0, 0, 1/2
    0, 0, 0
    0, 0, 0 ]

    A-I = E 1,1
    which is of course
    [ 0, 1, 1
    0, 0, 1
    0, 0, 0]

    and finally

    I = E1,0

    So we have 3 constituent matrices for A, let's say

    X E1,0 + Y E 1,1 + Z E 1,2

    It turns out for values X=1, Y= 1/2, and Z = -1/4 you get

    [ 1, 1/2, 3/8
    0, 1, 1/2
    0, 0, 1]

    whose square is A. So somehow (I don't know) we have to use the constituent matrices in a linear equation to general the square root of A. How to get the values of X,Y,Z I do not know.
    finding a square root of a 3 \times 3 upper triangular matrix A=[a_{ij}] is not hard (i'll assume that the entries on the diagonal of A are real and positive). define the 3 \times 3 matrix B=[b_{ij}] by:

    for i=1,2,3 let b_{ii}=\sqrt{a_{ii}}. also define b_{21}=b_{31}=b_{32}=0. finally let b_{12}=\frac{a_{12}}{b_{11} + b_{22}}, \ b_{23}=\frac{a_{23}}{b_{22}+b_{33}} and b_{13}=\frac{a_{13} - b_{12}b_{23}}{b_{11}+b_{33}}. it's easy to see that B^2=A.

    of course this formula will also work for matrices with complex entries if you choose a square root for each a_{ii} and if the denominators in b_{12}, \ b_{23} and b_{13} are non-zero. finally, since every

    square matrix with complex entries is similar to an upper triangular matrix, you can use the above to find a square root of an arbitrary 3 \times 3 matrix.
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  3. #3
    Newbie
    Joined
    Dec 2009
    Posts
    2
    Thank you very much.

    Is there a way to then determine the Jordan form of A?

    If eigenvalues are 1, then J=

    [ 1, 0, 0
    0, 1, 0
    0, 0, 1]
    or
    [1,1,0
    0,1,0
    0,0,1]
    or
    [1,1,0
    0,1,1
    0,0,1]
    or
    [1,0,0
    0,1,1
    0,0,1]

    Can we determine which is the Jordan form without finding P st A = (P^-1) (J) (P) ?
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