Eigendecomposing real matrices without knowing complex numbers Let $A$ be a real squared matrix. A scalar $\lambda\in\mathbb C$ is an eigenvalue for $A$ iff
$$Av=\lambda v \tag A$$
for some complex vector $v$.
This condition can be equivalently written in terms of purely real quantities as the following system:
$$\begin{cases}
  (A-\lambda_1 I)v_1 = - \lambda_2 v_2, \\
  (A-\lambda_1 I)v_2 = \phantom{-}\lambda_2 v_1,
\end{cases} \tag B$$
as can be seen by decomposing the quantities in (A) into real and imaginary parts: $\lambda=\lambda_1+i\lambda_2$ and $v=v_1+i v_2$.
If we didn't know anything about complex numbers, we would be working directly on (B), asking for a pair of reals $\lambda_1,\lambda_2\in\mathbb R$ such that (B) is satisfied for some real vectors $v_1,v_2$.
This pair of conditions can be seen to imply to the following ones:
$$\begin{cases}
  [(A-\lambda_1 I)^2 + \lambda_2^2 I ]v_1 = 0, \\
  [(A-\lambda_1 I)^2 + \lambda_2^2 I ]v_2 = 0.
\end{cases} \tag C$$
This follows from applying $(A-\lambda_1 I)$ twice to either $v_1$ or $v_2$, and using (B).
This, on the other hand, is equivalent to the condition
$$\det[(A-\lambda_1 I)^2 + \lambda_2^2 I] = 0. \tag D$$
See also this post about the equivalence of (A) and (D).
From complex analysis we know that, given an arbitrary real matrix $A$, there must be a pair of reals $\lambda_1,\lambda_2$ such that (D) is verified.
Not knowing what complex numbers are, how would we go in finding such values for a given $A$? The determinant equation gives a polynomial of two variables which I'm not sure how to handle.
 A: The $\lambda$-eigenspace of an operator is associated to the linear polynomial $p(x) = x - \lambda$:


*

*The $\lambda$-eigenspace of $A$ is the set $\{v \mid p(A)v = 0\}$.

*The presence of $(x - \lambda)$ in the characteristic polynomial of $A$ tells you that there is a $\lambda$-eigenspace.


For example, if the characteristic polynomial of $A$ is $x^2 - 2x - 3$, then I can factorise it to $(x-3)(x+1)$ and hence I know that the two eigenvalues are $3$ and $-1$.
However, over the real numbers not every real polynomial factorises into linear parts: the best we can do is linear parts and quadratic parts. Say for example that the characteristic polynomial of $A$ was $(x-1)(x^2 + 1)$. Then the $(x-1)$ term tells me that there is a one-dimensional subspace $L$ such that $(A-1)L = 0$ (i.e. a 1-eigenspace), while the $(x^2 + 1)$ part tells me that there is a two-dimensional subspace $P$ such that $(A^2 + 1)P = 0$. 
For each irreducible quadratic you see, there is a pair of complex conjugates associated to it (the roots). This pair of complex conjugates are your $\lambda_1 \pm i\lambda_2$. For example, for the polynomial $x^2 + 1$, the complex conjugates would be $\pm i$.
If you for some reason absolutely didn't want to touch complex numbers, and you had an irreducible quadratic $ax^2 + bx + c = 0$ that you were trying to extract $\lambda_1$ and $\lambda_2$ from, you could just break up the quadratic formula into real and imaginary parts:
$$\lambda_1 = \frac{-b}{2a}, \quad \lambda_2 = \pm \frac{\sqrt{4ac - b^2}}{2a}$$
This is not really a cheat, since we're just completing the square on $ax^2 + bx + c$ to make it look more like $(x-\lambda_1)^2 + \lambda_2^2$.
