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Raising Matrices to power of n. (complex)

  1. Dec 18, 2008 #1
    So lets say you have a matrice that has a det of 0

    So X= [1 1; 1 1]

    So using some verification and algebra u arive at

    X^n= 2^(n-1)[1 1; 1 1]

    So I am trying to define what n can be.

    I already established n cannot be a - since one would get infinity for an answer since it is a singular matrix.

    Another limiation is 0. Since that just gives you an identity which the formula cannot express.

    But I am confused with fractions.

    Technically they should work but I do not know how to explain it.
    Same with irrational numbers.

    so can n=1/2?

    and if that is not hard enough what about imaginary numbers?????????

    For some reason I know they work with non singular matrices .
  2. jcsd
  3. Dec 19, 2008 #2
  4. Dec 19, 2008 #3
    Your formula does work for n=1/2. (If you square 2^(1/2-1) [1,1;1,1] you get back your original matrix.) In fact, the formula works for n=1/b for any positive integer b. Since you know it works for positive integers it also works for n=a/b for positive integers a/b. So it works for positive rational numbers.

    For irrational and imaginary n you have to decide what it means to raise a matrix to this type of power. Hopefully someone here will have an idea.

    Mathematica actually lets you raise your matrix to a positive irrational power and it gives back your formula. (It gave an error message when I tried to do imaginary powers). I don't know how to interpret this result, though.
  5. Dec 19, 2008 #4

    Ben Niehoff

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    For imaginary (or in general, complex) powers, the definition for ordinary numbers (not matrices) is

    [tex]\alpha^{\beta} = e^{\beta \ln \alpha}[/tex]

    For a matrix A, this is only defined if ln(A) exists. This is not always the case (in fact, I think det(A)=0 is precisely one case for which ln(A) is not defined...the determinant is something like the "magnitude" of a matrix).

    Note: One can take arbitrary functions of matrices f(A), so long as f(A) can be defined as a convergent Taylor series. The matrix logarithm can be given by

    [tex]\ln (1+A) = A - \frac12 A^2 + \frac13 A^3 - \frac14 A^4 + ...[/tex]

    which does not converge for your matrix.
    Last edited: Dec 19, 2008
  6. Dec 19, 2008 #5
    I used matlab and it gave values for fractions. So I am confused there.

    Is there a proof for fractions?

    I want to algebriacly proove whyy fractions work. Perhaps through induction?
  7. Dec 19, 2008 #6


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    A proof of what for fractions? You haven't stated what it is you want to prove.

    The matrix
    [tex]A= \left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right][/tex]
    has two distinct eigenvalues and so is diagonalizable. That is, there exist a matrix P such that [itex]P^{-1}AP= D[/itex] where D is a diagonal matrix having the eigenvalues 0 and 1 on the main diagonal. Then [itex]A= PDP^{-1}[/itex] and
    [tex]A^n= (PDP^{-1})(PDP^{-1})(PDP^{-1})\cdot\cdot\cdot(PDP^{-1})= PD^nP^{-1}[/tex]
    where Dn is easy- it is the diagonal matrix with 0 and 2n on the main diagonal.

    Specifically, an eigenvector corresponding to the eigenvalue 0 is <1, -1> and an eigenvector corresponding to the eigenvalue 1 is <1, 1> so we can take
    [tex]P= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right][/tex]
    and then
    [tex]P^{-1}= \left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right][/tex]

    That is
    [tex]\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}0 & 0 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right][/tex]

    and so
    [tex]\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]^n= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}0 & 0 \\ 0 & 2^n\end{array}\right]\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right][/tex]

    You can show that the formula works for 1/n as well by reversing it:
    [tex]\left[\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right]= \left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{cc}0 & 0 \\ 0 & 2^{1/n}\end{array}\right]^n\left[\begin{array}{cc} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right][/tex]
  8. Dec 19, 2008 #7
    ). I basicly want to prove that when you have a singular matrix X. And you raise it to the power of 0<n<1 one can yield results.

    Thanks for the exmplanation I get how to do it now. But I need to research more on eigenvalues and eigenvectors since in the course I am in currently we do not learn about those key properties.

    Couple more things:
    Without using eigen values can you simply prove that by saying
    Where M is the result?

    and I gota questions can you also use taylor series to explain why irrational numbers work?

    Or perhaps like a pi expansion where you put fractions for n for your required precesion?

    An why exactly doesnt ln of X work when logs work.
    Last edited: Dec 19, 2008
  9. Dec 20, 2008 #8


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    If X1/n= M then Mn= X, not "X1/n= Mn".

    I don't know what you mean by "why exactly doesnt ln of X work when logs work. "

    ln (the natural logarithm) works whenever any log "works". loga(X)= ln(X)/ln(a).
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