Let $A$ be a $2 \times 2$ real matrix such that $A^2 - A + (1/2)I = 0$. Prove that $A^n \to 0$ as $n \to \infty$. Question: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^2 - A + (1/2)I = 0$, where $I$ is the $2 \times 2$ identity matrix and $0$ is the $2 \times 2$ zero matrix. Prove that $A^n \to 0$ as $n \to \infty$.

My attempt: Here is my idea so far. Consider $A$ as a $2 \times 2$ matrix over the field of complex numbers. Now, the polynomial 

$$g(t) = t^2 - t + \frac{1}{2} $$

factors as 

$$g(t) = \left(\frac{1}{2} - \frac{i}{2}\right)\left(\frac{1}{2}+\frac{i}{2}\right)$$

over $C$. This means that the minimal polynomial of $A$ over $C$ is: 

$$m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right), m(t) = \left(t-(\frac{1}{2} + \frac{i}{2})\right), \ \ \ \text{ or } \ \ \ m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right)\left(t-(\frac{1}{2} + \frac{i}{2})\right),$$

and thus we have $A = Q D Q^{-1}$, where

$$D =\left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&amp;0\\
0&amp;{\frac{1}{2}(1 - i)}
\end{array}} \right],$$

$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 + i)}&amp;0\\
0&amp;{\frac{1}{2}(1 + i)}
\end{array}} \right],\ \ \ \text{ or } $$

$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&amp;0\\
0&amp;{\frac{1}{2}(1 + i)}
\end{array}} \right].$$

Now, $A^n = QD^nQ^{-1}$. 

Here is where I get stuck. $D^n$ doesn't seem to be converging to $0$ as $n$ approaches infinity. In addition, I am concerned that my strategy is bad, since we are talking about a real matrix and I'm using a minimal polynomial over $C$. Is it still true that $A$ must have one of the three forms above, even if $A$ is supposed to be real? 

Thanks for any help/suggestions you may be able to provide.
The eigenvalues of $A$ (or $D$) satisfy $|\lambda | = { 1\over \sqrt{2}} &lt;1$. Hence
$|\lambda|^n \to 0$.
For generalisation, assume $$
        A=\begin{bmatrix}
        a &amp; b \\
        c &amp; d \\
        \end{bmatrix}
$$ Now, for every $2X2$ matrix, this equation is always valid $A^2-trace(A)A+det(A)I=0$ (prove this by youself, this is very easy). Now you can compare this equation with your original equation, you will get $a+d=-1$ and $ad-bc=\frac {1}{2}$. Assume any values of $a,b,c,d$ satistfying these equations which I had done earlier.

Assume the values $a=d=-0.5, -b=c=0.5$, then matrix will look like $$2A=
        \begin{bmatrix}
        -1 &amp; -1  \\
         1 &amp; -1  \\
        \end{bmatrix}
$$ This matrix satisfy all the given condition.Now you can proceed. You can get very easily that $A^n=0$ as $n \to \infty$ . And also you can also use the eigen value logic to prove that $A^n=0$ as $n \to \infty$

Eigen Value Logic can also be applied. Let the eigen value of this matrix be $e_1$ and $e_2$. From given equation it is clear that $det(A)=0.5=e_1e_2$ and both eigen values are non real. So $e_1e_2=e^2=0.5$ where $|e_1|=|e_2|=e$

Now $\alpha=|A^n|=|A|^n=2^{-n}$ and as $n \to \infty$, $\alpha=0$ which clearly proves that $A^n$ has to be zero matrix because determinant of that matrix is zero, otherwise determinat of that matrix will never be zero.