Unclear problem with $n$-th power matrix and limit

Question

Find $$\lim\limits_{n \to \infty} \frac{A_n}{D_n}$$ where $$\begin{pmatrix} 19 & -48 \\ 8 & -21 \\ \end{pmatrix} ^{\! n} = \begin{pmatrix} A_n & B_n \\ C_n & D_n \\ \end{pmatrix}$$

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$n$ - is the power of a matrix, but what is $A_n, B_n, C_n, D_n$ then? Is it a corresponding element of a matrix in the $n$-th power? How is this type of problem called? And what is the way to solve that problem?

Answer 1

$$A:=\begin{pmatrix}19&-48\\8&-21\end{pmatrix}=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}-5&0\\0&3\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=PDP^{-1}$$

So $A^n=PDP^{-1}PDP^{-1}\cdots PDP^{-1}=PD^nP^{-1}$

That is, $$A^n=\begin{pmatrix}2&3\\1&1\end{pmatrix}\begin{pmatrix}(-5)^n&0\\0&3^n\end{pmatrix}\begin{pmatrix}2&3\\1&1\end{pmatrix}^{-1}=\begin{pmatrix}-2 (-5)^n + 3^{1 + n}& 6 (-5)^n - 2\times 3^{1 + n}\\-(-5)^n + 3^n& 3 (-5)^n - 2\times 3^n\end{pmatrix}$$

Answer 2

Here is an unconventional approach: we have $$ M = \pmatrix{19 & -48\\ 8 & -21}. $$ We find that the eigenvalues satisfy $$ \det(M - xI) = x^2 + 2x - 15 = 0 \implies x = -5,3. $$ By the Cayley Hamilton theorem, the powers of $M$ satisfy the recurrence $$ M^n + 2M^{n-1} -15 M^{n-2} = 0 $$ From the theory of constant coefficient homogeneous linear difference equations, it follows that $M^n$ has the form $$ M^n = (-5)^n P + 3^n Q $$ for some matrices $P,Q$. We can solve for $P,Q$ using the "initial conditions" of $n=0,1$. We have $$ P + Q = M^0 = \pmatrix{1&0\\0&1}, \quad (-5)P + 3Q = M^1 = \pmatrix{19 & -48\\8 & -21}. $$ Subtracting the second equation from $3$ times the first yields $$ 3P - (-5)P + 0Q = 3\pmatrix{1&0\\0&1} - \pmatrix{19 & -48\\8 & -21} \implies\\ 8P = \pmatrix{-16 & -48\\8 & 24} \implies P = \pmatrix{-2&-6\\1&3}. $$ We use the first equation to find $$ Q = \pmatrix{1&0\\0&1} - P = \pmatrix{3&6\\-1&-2}. $$ So, we have $$ M^n = (-5)^n \pmatrix{-2&-6\\1&3} + 3^n\pmatrix{3&6\\-1&-2}. $$