If $A=\pmatrix{1 &0\\-1&1}$, show that $A^2-2A+I_2=0$. Hence find $A^{50}$ 
If $$A=\pmatrix{1 &0\\-1&1},$$ show that $$A^2-2A+I_2=0,$$ where $I_{2}$ is the $2x2$ Identity matrix. Hence find $A^{50}$.  

We have $$A^2-2A+I_2=A(A-2I_3)+I_=\pmatrix{1 &0\\-1&1}\pmatrix{-1 &0\\-1&-1}+I_2  
=-I_2+I_2=0.$$
How can I show the second part?
 A: Since $(A-I_2)^2 = 0$, by binomial theorem, one has
$$\begin{align}A^{50}
&=  (I_2 + (A - I_2))^{50}\\
&= I_2 + \binom{50}{1} (A-I_2)^{1} + \binom{50}{2}(A-I_2)^2 + \binom{50}{3}(A-I_2)^3 +\cdots\\
&= I_2 + 50 (A-I_2) + 0 + 0 + \cdots\\
&= \begin{pmatrix}1 & 0\\-50 & 1\end{pmatrix}
\end{align}
$$
A: The characteristic polynomial of $A$ is
$$\chi_A(t) = \text{det}(A-t\Bbb{1})=t^2-2t+1$$
By the theorem of Caliley Hamilton, you get that $A^2-2A+\Bbb{1}=0$, where $\Bbb{1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $0$ is the zero matrix.
Now we 

Claim $$A^n = \begin{pmatrix}1&0\\-n&1\end{pmatrix}$$

Proof
We procede by induction:
If $n=1$ we have $A = \begin{pmatrix}1&0\\-1&0\end{pmatrix}$, that is correct.
Assume that the claim is true for $n-1$. We want to prove that it holds also for $n$
\begin{align*}
A^n &= A^{n-1}\cdot A\\
&=\begin{pmatrix}1&0\\-(n-1)&1\end{pmatrix}\cdot \begin{pmatrix}1&0\\-1&1\end{pmatrix}\\&=\begin{pmatrix}1&0\\-n+1-1&1\end{pmatrix}=\begin{pmatrix}1&0\\-n&1\end{pmatrix}
\end{align*}
Which proves the claim. Now you should be able to evaluate $A^{50}$.
A: You've established that $A^2 -2A + I = 0$
Rearrange, $A^2 = 2A - I$
Let's compute $A^3$:
$A^3 = A^2 A = 2A^2 - IA = 2(2A - I) - A = 3A - 2I$
Similarly, confirm that $A^4 = 4A - 3I$
We can now conjecture that $A^n = nA - (n-1)I, n \geq 2$
We can prove that by induction.
The base case is obvious (work already done for $n=2$).
Assume it's true for some $n=k$, i.e. $A^k = kA - (k-1)I$
Hence $A^{k+1} = A^kA = kA^2 - (k-1)IA = k(2A-I) - (k-1)A = 2kA - kI - kA + A = (k+1)A - kI$
which completes the proof.
Therefore $A^{50} = 50A - 49I = 50\pmatrix{1 &0\\-1&1} - 49\pmatrix{1 &0\\0&1} = \pmatrix{1 &0\\-50&1}$.
