How would I go about calculating:
$ \left( \begin{array}{cc} 3 & -\sqrt{3} \\ \sqrt{3} & -1 \end{array} \right)^{13}$
I have already attempted to find eigenvalues/eigenvectors, but I believe I am missing a much simpler method.
$\endgroup$4 Answers
$\begingroup$Evaluate the eigenvalues and also the matrix P and D so $ A=PDP^{-1} $
then $ A^{n} =PD^{n}P^{-1} $
since the matrix D is diagonal then $ D^{n} $ has the element
$$ a12=a21=0 $$ and $ a11= \lambda _{1}^{n} $
$ a22= \lambda _{2}^{n} $
here lambda are the 2 eigenvalues
$\endgroup$ $\begingroup$Another approach you may use is to break $13$ down into powers of two. We see that $13=8+4+1$. You can compute $A^8$, $A^4$ and then find your desired product by $A^8 \cdot A^4 \cdot A = A^{13}$.
Powers of two are easier to handle than other powers, and so this method is used often when computing large powers.
This is because $A^2=A\cdot A$, $A^4 = A^2 \cdot A^2$, and $A^8=A^4 \cdot A^4$. Therefore three multiplications are needed to find $A^8$ and $A^4$. Then two more multiplications are needed to compute $A^{13} = A^8 \cdot A^4 \cdot A$. Thus only five multiplications are needed. This is in contrast to the necessary 12 products that need to be computed in a direct method.
$\endgroup$ $\begingroup$Hint: Have you tried multiplying your matrix by itself $M^2=?$
$\endgroup$ $\begingroup$Try finding $M^2$ and $M^3$. Observe a pattern and conjecture $(M^n)_i= 2(M^{n-1})_i$ for $i=1,2,3,4$. Prove the conjecture by induction.
Thus, we obtain $(M^n)_i=2^{n-1} (M_1)_i$.
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