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Thread: Raising Matrices to power of n

  1. #1
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    Raising Matrices to power of n

    $\displaystyle A=\begin{pmatrix}a & 0 & b\\ 0 & a+b & 0\\ b & 0 & a\end{pmatrix}$

    Um, well, I know I should find a formula for $\displaystyle A^n$ and prove it using induction. I calculated the first four powers of A, but still didn't observe any rule >.>

    $\displaystyle A^2=\begin{pmatrix}a^2+b^2 & 0 & 2ab\\ 0 & (a+b)^2 & 0\\2ab & 0 & a^2+b^2\end{pmatrix}$

    $\displaystyle A^3=\begin{pmatrix}a^3+3ab^2 & 0 & b^3+3a^2b\\ 0 & (a+b)^3 & 0\\b^3+3a^2b & 0 & a^3+3ab^2\end{pmatrix}$

    $\displaystyle A^3=\begin{pmatrix}a^3+3ab^2 & 0 & b^3+3a^2b\\ 0 & (a+b)^3 & 0\\b^3+3a^2b & 0 & a^3+3ab^2\end{pmatrix}$

    $\displaystyle A^4=\begin{pmatrix}a^4+b^4+6a^2b^2 & 0 & 4a^3b+4ab^3\\ 0 & (a+b)^3 & 0\\4a^3b+4ab^3 & 0 & a^4+b^4+6a^2b^2\end{pmatrix}$

    Actually, it's not really nothing, but still useless:
    $\displaystyle a^2+b^2+2ab=(a+b)^2$
    $\displaystyle a^3+3ab^2+b^3+3a^2b =(a+b)^3$
    $\displaystyle a^4+b^4+6a^2b^2+4a^3b+4ab^3=(a+b)^4$

    Thanks in advance for any ideas! ^^
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  2. #2
    Super Member girdav's Avatar
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    Re: Raising Matrices to power of n

    Note that $\displaystyle A=aI_3+bJ$ where $\displaystyle J=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}$.
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  3. #3
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    Re: Raising Matrices to power of n

    Mhm, but $\displaystyle J^n$ have two different forms for n even and odd. So that doesn't help me much, I mean after using the binomial theorem I'll have a... uhm, ugly sum.
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  4. #4
    Super Member girdav's Avatar
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    Re: Raising Matrices to power of n

    You can write $\displaystyle J^k=\frac{1+(-1)^k}2I +\frac{1-(-1)^k}2J$. You will get a result in terms of $\displaystyle (a+b)^n$, $\displaystyle (a-b)^n$, $\displaystyle I$ and $\displaystyle J$.
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