How can I find a $2x2$ matrix with no real eigenvalues? I am trying to see if there is a process to finding a matrix with no real eigenvalues.
I know when we get to the point of $\lambda^{2} + 1 = 0$ then this will have no real solution.
Is there a way to work "backwards" and find a matrix, or is this just intuition? 
 A: This is a correct way of proceeding. A matrix of the following form
$$A = \begin{bmatrix}-\lambda & 1 \\ -1 & - \lambda\end{bmatrix}$$
has $\det(A) = \lambda^2 + 1$. What is $B$ if $A = B - \lambda I$?
A: One can construct a matrix $A\in\mathbb{R}^{2x2}$,
$$A = \begin{pmatrix}
a & -b\\
b & a
\end{pmatrix},$$
which has characteristic polynomial $p_A(t) = (t-a)^2+b^2$. This form produces every possible conjugate pair of complex solutions.
A: One can proceed naively and compute the characteristic polynomial $c_A$ of a general matrix $$A = \pmatrix{a&b \\ c&d}$$
whose roots, as you know, are the eigenvalues of $A$:
$$\begin{align*}
c_A(\lambda) = \det(\lambda I - A) &= \det \pmatrix{\lambda - a & -b \\ -c & \lambda - d} \\
&= (\lambda - a)(\lambda - d) - (-b) (-c) \\
&= \lambda^2 - (a + d) \lambda + (a d - b c)
\end{align*}$$
Now, $c_A$ is a quadratic function of $\lambda$, so it has no real roots---equivalently, $A$ has no real eigenvalues---if its discriminant $\Delta$ is negative:
$$
    \Delta = [-(a + d)]^2 - 4 (1) (a d - b c) = (a - d)^2 + 4 b c .
$$
So, we need only pick $a, b, c, d$ to make this quantity negative.
For example, taking $a = d, c = -b$---which up to renaming of variables recovers the examples given in the other two answer---gives $\Delta = - 4 b^2$, which is negative, and so determines a matrix without real eigenvalues, provided that $b \neq 0$
