Let $A$ be a $5\times5$ matrix, such that any element $a_{ij}$ equals either 1 or -1. Find all solutions of the equation: $$AX = \frac{1}{2}X$$ where $X \in \mathbb{R}^5$.
What I've tried:
If $X$ is some solution, then $X' =(\alpha X - \beta X)$ is also a solution ($\alpha , \beta \in \mathbb{R}$).
This means, that we can get such solution $X$ (if any exists at all), that $x_1 = 1$. This simplifies a bit our system of linear equations:
$$\begin{cases} \pm 1 \pm x_2 ... \pm x_5 = \frac{1}{2} \\ ... \\ \pm 1 \pm x_2 + ... + \pm x_5 = \frac{1}{2}x_5 \end{cases}$$
Here I tried to solve this system, but got a mess of "$\pm$" signs and think that this is not the way this task should be solved.
The characteristic polynomial is of the form $$t^5-a_1 t^4 + a_2 t^3 - a_3 t^2 +a_4 t -a_5,$$ where the $a_j$ are integers. Suppose $\frac{1}{2}$ is a characteristic root. Then substitute in, multiply by 32 and --- 1 is even!
So we see that the only possible $X$ is $0$.