Show that $\lim_{n\rightarrow\infty} A ^{n} = B$ Lin alg final coming up, getting the study sesh in
Fun fact:just learned latex so yay
The full problem comes in two parts. Part A is simple/
Given $A$=  \begin{pmatrix}
2 & -3/2  \\
1 & -1/2  \\
 \end{pmatrix} 
and
$B$= \begin{pmatrix}
3 & -3  \\
2 & -2  \\
 \end{pmatrix} 
a. Find a matrix $Q$ such that $Q^{-1}AQ= 
D$ where $D$ is a diagonal matrix.
Which is simple enough, found the eigen-vectors that came out to be $Q$:
 \begin{pmatrix}
1 & 3/2  \\
1 & 1  \\
 \end{pmatrix}
(which could very well be wrong).
Part B is where the difficulty arises:
b. Show that $\lim_{n\rightarrow\infty}  A ^{n} =  B$ 
I'm assuming this has to be solved using some eigen-vectors or some orthogonal basis transformation, however I cannot seem to wrap my head around this problem.
Thank you in advance!
PS. Thanks to John Moeller for the formatting tip.
 A: If $A = Q^{-1}DQ$, then $$
  A^n = \underbrace{(Q^{-1}DQ)(Q^{-1}DQ) \ldots (Q^{-1}DQ)}_{n\text{ times}} = Q^{-1}D^nQ \text{,}
$$
and if $D = \textrm{diag}(d_1,\ldots,d_k)$ then $D^k = \textrm{diag }(d_1^n,\ldots,d_k^n)$.
It follows that for $A = Q^{-1}DQ$ and $D = \textrm{diag }(d_1,\ldots,d_k)$ $$\begin{eqnarray}
 (1) && \lim_{n\to\infty} A^n \text{ exists if } d_i \in (-1,1] \text{ for all } i,
 &\text{ and in this case} \\
 (2) && \lim_{n\to\infty} A^n = Q^{-1}\hat{D}Q \quad \text{where}
 & \hat D = \textrm{diag }(\hat d_1,\ldots,\hat d_k),\, \\
 &&& \hat d_i = \begin{cases}
    1 &\text{if $d_i=1$} \\
    0 &\text{if $-1 < d_i  < 1$.}
  \end{cases}
\end{eqnarray}$$
Note that the conditions on the $d_i$ simply guarantee that $\lim_{n\to\infty} d_i$ exists, and $\hat d_i$ is then defined so that $\hat d_i = \lim_{n\to\infty} d_i$. Also note that the $d_i$ are exactly the eigenvalues of $A$. 
It follows that, at least for a diagonalizable $A$, $\lim_{n\to\infty} A^n$ exists exactly if all the eigenvalues of $A$ lie within $(-1,1]$.
A: Hint: Since you know that $Q$ is invertible, you know that you can set $A = QDQ^{-1}$, right? So what's $A^2$? $A^3$? $A^n$?
A: The matrix $A$ will have eigenvalues $1$ and $\frac{1}{2}$, so
$$A^n=Q\pmatrix{1&0\cr0&({\textstyle\frac{1}{2}})^n\cr}Q^{-1}\ .$$
Can you see what you get for the bit in the middle as $n\to\infty$?  If you can do this, then since you have already found $Q$, the problem is solved.
