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Matrix is a perfect square?

  1. Aug 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Is the following matrix a perfect square? A matrix A is a perfect square if there is a matrix B such that A = B2

    ____1 1 0 0
    A = 0 1 0 0
    -----0 0 1 0
    -----0 0 0 1

    2. Relevant equations



    3. The attempt at a solution

    I haven't come across anything like this before, but here's what I'm thinking.

    All eigenvalues of A are equal to 1, and the minimal polynomial is m(t) = (t-1)2
    and the characteristic polynomial is p(t) = (t-1)4.


    Both of these are perfect squares, so does that mean the matrix is a perfect square?
     
  2. jcsd
  3. Aug 31, 2011 #2

    Ray Vickson

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    Here is a hint: look at the function f(A) = sqrt(e*I + A), where I = identity matrix and e > 0 is a small parameter. You can compute f(A) from the fact that A is a Jordan form with one Jordan block of size 2 and two of size 1, and all eigenvalues are 1. Basically, if f(x) = sum c_n*x^n is an analytic function, then f(A) = c_0*I + sum_{n=1..infinity} c_n * A^n; I = A^0. Since A is so simple, you can get all the A^n easily and do the sum. After you have a nice final formula for f(A) you can see if it has a definite limit as e --> 0. Or, for a quicker way see, eg, Gantmacher, Theory of Matrices, or Google "analytic functions of matrices".

    RGV
     
    Last edited: Aug 31, 2011
  4. Aug 31, 2011 #3

    I like Serena

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    Hi Maybe_Memorie! :smile:

    What you're thinking may be true or not.
    I do not know (yet).
    But without proof you cannot use it.

    Here's another method.
    If A is a perfect square, it should have one or more square roots.
    One method to find a square root is to calculate a power series expansion for a broken power:
    [tex](I+X)^{1 \over 2}=I + {1 \over 2} X - {1 \over 8} X^2 + ...[/tex]
    Are you already familar with those?
     
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