Originally Posted by

**Opalg** First, check what you have done so far. (I think that the eigenvalues should be 3, 6, 9, not 3, –6, 9.)

Next, look through those old notes to see that the normalised eigenvectors form the columns of an orthogonal matrix P with the property the $\displaystyle A = PDP^{-1},$ where D is the diagonal matrix whose diagonal entries are the eigenvalues. Then $\displaystyle A^n = PD^nP^{-1}.$

So to find $\displaystyle A^n$ it is sufficient to find $\displaystyle D^n$. But if $\displaystyle D = \begin{bmatrix}3&0&0\\0&6&0\\0&0&9\end{bmatrix}$ then $\displaystyle D^n = \begin{bmatrix}3^n&0&0\\0&6^n&0\\0&0&9^n \end{bmatrix}$. So the question is asking you to look at the limit as n goes to infinity of $\displaystyle \tfrac1{c^n}D^n = \begin{bmatrix}\tfrac{3^n}{c^n}&0&0\\0&\tfrac{6^n} {c^n}&0\\0&0&\tfrac{9^n}{c^n}\end{bmatrix},$ for all positive values of c.

If c<9 then the limit will not exist, because at least one of the fractions in that matrix will go to infinity. If c>9 then the limit is the zero matrix. Finally, the only interesting case occurs when c=9, in which case $\displaystyle \lim_{n\to\infty}D^n = \begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix},$ and $\displaystyle \lim_{n\to\infty}A^n = P\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}P^ {-1}.$