Find eigenvalues and eigenvector analitically Let's consider a matrix $M$ so that
$$
M = \begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
$$
We want to find the eigenvalues and eigenvectors of that matrix so, as the rule says, we have to solve
$$ \det(M - \lambda I) = 0 $$
Developing this equation gives us something of the form
$$ A\lambda^2 + B\lambda + C = 0 $$
We know that this equation admits either two real solutions, or one real solution or two imaginary solutions.
Q1: what does mean having two imaginary solutions in the case of eigenvalues?
Q2: what does imply having a single solution? Is this even possible ?
Let $ \lambda_i $ be the i-th solution to the above equation. In order to find the eigenvector $v_i = [v_{i1}, v_{i2}]^T$ associated to $\lambda_i$, we have to solve
$$ (M - \lambda_i I)v_i = 0 $$
For each eigenvalue, this gives a system of two equations with two unknowns:
$$\begin{equation}\begin{aligned}
(a - \lambda_i)v_{i1} + bv_{i2} = 0 (Eq1) \\
cv_{i1} + (d - \lambda_i)v_{i2} = 0 (Eq2) \\
\end{aligned}\end{equation}$$
In order to solve this, I know of two methods:

*

*substract: $(Eq1)*c - (Eq2)*(a - \lambda_i)$
$$\begin{equation}\begin{aligned}
c(a - \lambda_i)v_{i1} + cbv_{i2} - (a - \lambda_i)cv_{i1} - (a - \lambda_i)(d - \lambda_i)v_{i2} = 0 \\
cbv_{i2} - (a - \lambda_i)(d - \lambda_i)v_{i2} = 0 \\
\left ( cb - (a - \lambda_i)(d - \lambda_i) \right ) v_{i2} = 0 \\
\end{aligned}\end{equation}$$
The only solution I see is $v_{i2} = 0$ which leads to $v_{i1} = 0$ and is not what I expect..


*substitution

We can re-write $v_{i1} $ from equation (Eq1) with
$$ v_{i1} = -\frac{b}{(a - \lambda_i)}v_{i2} $$
Which gives in equation (Eq2)
$$\begin{equation}\begin{aligned}
-c\frac{b}{(a - \lambda_i)}v_{i2} + (d - \lambda_i)v_{i2} = 0 \\
\left( (d - \lambda_i) - \frac{b}{(a - \lambda_i)} \right ) v_{i2} = 0 \\
\end{aligned}\end{equation}$$
The only solution I see is $v_{i2} = 0$ which leads to $v_{i1} = 0$ and is not what I expect either..


*Lets consider the following practical example:

$$
M = \begin{bmatrix}
2 & 1 \\
1 & 2 \\
\end{bmatrix}
$$
It accepts $\lambda_1 = 1$ and $\lambda_2 = 3$ as eigenvalues.
Now going back to equations to find out their associated eigenvectors, we are supposed to find $v_1 = [ 1, -1 ]^T $ for $\lambda_1 = 1$, and $v_2 = [ 1, 1 ]^T$ for $\lambda_2 = 3$ ..
Q3: What is the right approach to this problem? Have I done anything wrong or misunderstood something at some point?
Q4: The provided example may be a particular case (?), are there some rules which may help decide the way to go for solving theses equations ?
 A: Thanks to @Dietrich and @Moo who commented with pieces of information which helped me get my way out!
Q1: Having eigenvalues being imaginary seems not to imply anything different
from having eigenvalues being real.
Q2: Having a single eigenvalue means that the diagonal matrix has multiple
time the same eigenvalue. And so it is for the eigenvectors.
Q3: The solution is to rewrite the equation in order to get a relation between the terms of the eigenvector, NOT A PARTICULAR VALUE.
In the case of a $2 \times 2$ matrix
$
M = \begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
$
And knowing that
$ (M -\lambda I)v = 0  <=>  Mv = \lambda v $
We get
$\begin{equation}\begin{aligned}
av_{i1} + bv_{i2} = \lambda v_{i1} \\
cv_{i1} + dv_{i2} = \lambda v_{i2} \\
\end{aligned}\end{equation}$
Which we solve by subtraction
$\begin{equation}\begin{aligned}
acv_{i1} + cbv_{i2} = c\lambda v_{i1} \\
acv_{i1} + adv_{i2} = a\lambda v_{i2} \\
\end{aligned}\end{equation}$
$\begin{equation}\begin{aligned}
(cb - ad)v_{i2} = \lambda cv_{i1} - \lambda av_{i2} \\
v_{i2} = \frac{c\lambda}{a\lambda + cb - ad} v_{i1} \\
\end{aligned}\end{equation}$
For the particular matrix provided as an example in the original question
$
M = \begin{bmatrix}
2 & 1 \\
1 & 2 \\
\end{bmatrix}
$
We obtain for $ \lambda_1 = 1$ the relation $ v_{12} = -v_{11}$ and for $ \lambda_2 = 3$ the relation $ v_{22} = v_{21}$.
Thus we can have a multitude of valid eigenvectors as long as the relation is satisfied. One set of them could be $v_1 = [ 1, -1 ]^T$ and $v_2 = [ 1, 1 ]^T$.
