# Thread: Raising Matrices to power of n

1. ## Raising Matrices to power of n

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

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

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

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

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

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

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

Thanks in advance for any ideas! ^^

2. ## Re: Raising Matrices to power of n

Note that $A=aI_3+bJ$ where $J=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}$.

3. ## Re: Raising Matrices to power of n

Mhm, but $J^n$ 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.

4. ## Re: Raising Matrices to power of n

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