Finding the eigenstates of an operator I am currently taking a course in QM and can't see how the eigenstates have been found for examples like this one:

Question
Let $\phi _1$ and $\phi _2$ be two normalised wavefunctions orthogonal onto each other. Let the action of the operator $\hat{A}$ on these states be:
$$\hat{A} \phi _1 = \phi _2$$
$$\hat{A} \phi _2 = \phi _1$$
Then the states $\Psi _1 = \phi _1 + \phi _2$ and $\Psi _2 = \phi _1 - \phi _2$ are eigenstates of $\hat{A}$ corresponding to the eigenvalues $+1$ and $-1$ respectively, that is:
$$\hat{A} \Psi _1 = \Psi _1$$
$$\hat{A} \Psi _2 = -\Psi _2$$

So I can see the states $\Psi _1$ and $\Psi _2$ satisfy the eigenfunction equation ($\hat{A}\Psi = \lambda \Psi$) with the given eigenvalues but I can't see how the states were derived, they seem to have been plucked from thin air but just work! Lots of QM questions seem to utilise states in the form $(\alpha \phi _1 \pm \beta \phi _2)$ but I'm not sure why?!
Appreciate any help!
 A: Assume $\phi_{1,2}$ form a basis. Consider the eigenvalue equation for $\hat{A}$, i.e. $\hat{A}\psi=\lambda\psi$. If we apply $\hat{A}$ again we get the equation $\hat{A}^2\psi=\lambda^2\psi$. But note from the definition of $\hat{A}$, i.e. its action on the basis, that $\hat{A}^2 =\text{Id}$. Thus the previous equation gives us 
$$
\lambda^2=1 \rightarrow \lambda=\pm1
$$
So we have found the eigen values pretty easily. The question remains as to how to find the eigenvectors. To do this, we begin by writing $ \phi_1 = \left( \begin{matrix}
1 \\\
0           
\end{matrix}\right)$ and $ \phi_2 = \left( \begin{matrix}
0 \\\
1           
\end{matrix}\right)$. A little thought leads us to conclude that we can represent $\hat{A}$ with the matrix
$$ \hat{A} = \left( \begin{matrix}
0 & 1 \\
1 & 0 \\ 
\end{matrix}\right)$$
For $\lambda = 1$ we want to solve $ \left(\left( \begin{matrix}
0 & 1 \\
1 & 0 \\ 
\end{matrix}\right)-\left( \begin{matrix}
1 & 0 \\
0 & 1 \\ 
\end{matrix}\right) \right) \left(\begin{matrix} a \\b\end{matrix} \right) = \left(\begin{matrix} 0 \\0\end{matrix} \right)$, or $\left( \begin{matrix}
-1 & 1 \\
1 & -1 \\ 
\end{matrix}\right)\left(\begin{matrix} a \\b\end{matrix} \right) = \left(\begin{matrix} 0 \\0\end{matrix} \right)$. This yields the system tells us that $a=b$ and we are left to choose the normalization, so we say $a=b=1$. Thus the eigenvector corresponding to $\lambda=1$ is $\left(\begin{matrix} 1 \\1\end{matrix} \right) = \left(\begin{matrix} 1 \\0\end{matrix} \right) + \left(\begin{matrix} 0 \\1\end{matrix} \right) = \phi_1 + \phi_2$. The eigenvector for $\lambda=-1$ follows similarly.
This is a standard example of a two level system, which @march points out in a comment is very common in physics. As Cosmas Zachos points out in another comment, the mystery vanishes when you realize you can write operators on two level systems as a two by two matrix, and find the eigenvectors in the usual way. 
A: gabe has already given an answer in terms of matrices and using idempotency. I shall exhibit a rather dry approach. 
Assuming that the $\phi$’s form a basis, then any vector can be expressed as $\alpha\phi_1+\beta\phi_2$. 
Now let $\psi=\alpha\phi_1+\beta\phi_2$ be an eigenstate of $A$. Then we have,
$A\psi=A(\alpha\phi_1+\beta\phi_2)$
$\implies \alpha A\phi_1+\beta A \phi_2$
$\implies \alpha\phi_2+\beta\phi_1=\lambda\alpha\phi_1+\lambda\beta\phi_2$
$\implies \lambda\alpha=\beta , \lambda\beta=\alpha$
$\implies \lambda^2\alpha=\alpha$
$\implies \lambda = \pm 1$
Plugging these back into $\alpha-\beta$ relation gives us $\alpha=\pm\beta$
Finally $\psi = \phi_1 \pm \phi_2$
