Finding $n^{th}$ Power of a matrix using diagonalization

Question

I came across this problem:
Let $M=\left[\begin{matrix}{1/2}& {1/4}\\0&1\end{matrix}\right]$ and $x=\left[\begin{matrix}3\\4\end{matrix}\right]$. Find $\lim\limits_{n\to\infty}M^nx$.

Given answer to this problem is $\left[\begin{matrix}2\\4\end{matrix}\right]$.

My approach to the problem is as follows:
$\underline{My Solution}:$ From the characteristic equation $|M-\lambda I|=0$, I got eigen values $\lambda=1,1/2$ and corresponding eigen-vectors as $\left[\begin{matrix}1\\2\end{matrix}\right],\left[\begin{matrix}0\\1\end{matrix}\right]$ respectively.

Then, the model matrix comes out to be $P=\left[\begin{matrix}1& 0\\2&1\end{matrix}\right]$ and diagonal matrix $D=\left[\begin{matrix}1& 0\\0&1/2\end{matrix}\right]$.

We know that, $M^n = P^{-1}D^nP$, where $P^{-1}=\left[\begin{matrix}1& 0\\-2&1\end{matrix}\right]$.

$\implies \lim\limits_{n\to\infty}M^n=\lim\limits_{n\to\infty}\left(\left[\begin{matrix}1& 0\\-2&1\end{matrix}\right]\left[\begin{matrix}1& 0\\0&1/2\end{matrix}\right]^n\left[\begin{matrix}1& 0\\2&1\end{matrix}\right]\right)$.

$\implies \lim\limits_{n\to\infty}M^n=\left[\begin{matrix}1& 0\\-2&1\end{matrix}\right]\left[\begin{matrix}1& 0\\0&\lim\limits_{n\to\infty}(1/2)^n\end{matrix}\right]\left[\begin{matrix}1& 0\\2&1\end{matrix}\right]$.

$\implies \lim\limits_{n\to\infty}M^n=\left[\begin{matrix}1& 0\\-2&1\end{matrix}\right]\left[\begin{matrix}1& 0\\0&0\end{matrix}\right]\left[\begin{matrix}1& 0\\2&1\end{matrix}\right]=\left[\begin{matrix}1& 0\\-2&0\end{matrix}\right]$.

So my answer comes out to be $\left[\begin{matrix}1& 0\\-2&0\end{matrix}\right]\left[\begin{matrix}3\\4\end{matrix}\right]=\left[\begin{matrix}3\\-6\end{matrix}\right]$.

I saw another question, which uses the same approach (diagonalizability), but here my answer does not match the given answer of this problem.
My Question is: Please tell, what is wrong with my approach. If there is another "easy" approach (like using Caley-Hamilton) to solve this, do let me know.

Answer 1

One of your eigenvector is wrong.

$$\left[\begin{matrix}{1/2}& {1/4}\\0&1\end{matrix}\right]\begin{bmatrix} 0 \\ 1\end{bmatrix}= \begin{bmatrix} \frac14 \\ 1\end{bmatrix}$$

which is not a multiple of $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

Since the matrix is a tringular matrix, the eigenvalues can be read off from the diagonal and hence there is no need to form the charactheristic equation.

Answer 2

As $M$ is upper-triangular, then so is $M^n$ for all $n$. How then could $\lim_{n\to\infty} M^n$ not be upper triangular? Looking at the diagonal, $N=\lim_{n\to\infty} M^n$ must be $\pmatrix{0&b\\0&1}$ for some $b$. As $MN=NM$ then $b=b/2+1/4$, that is $b=1/2$. Now you can work out $N\pmatrix{3\\4}$.