$v \neq 0$ is said to be an eigenvector of $M$ iff there exists $\lambda \neq 0 \in F$, such that \(M \cdot v = \lambda \cdot v\) and $\lambda$ is the eigenvalue of $v$.
Also note, that if we have $w = \lambda \cdot v$ then
\(\begin{align}
M \cdot w &= (M \lambda) \cdot v \\
&= (\lambda M) \cdot v \\
&= \lambda (M \cdot v) \\
&= \lambda \lambda \cdot v \\
&= \lambda^2 \cdot v \\
\end{align}\)
Thus, $w$ is also an eigenvector.
Similarly, we can see that $M^t \cdot v = \lambda^t \cdot v$.
Given a linear transform $M$, let us suppose there exists some eigenvector $v$.
Then,
\(\begin{align}
M \cdot v &= \lambda \cdot v \\
\implies M \cdot v - \lambda I_n \cdot v &= 0 \\
\implies (M - \lambda I_n ) \cdot v &= 0 \\
\implies (M - \lambda I_n ) \cdot v &= 0 \\
\end{align}\)
So this means that the vector $v$ belongs to the null space of the transformation $M - \lambda I_n$.
However note that we explicitly took $v$ to be a non-zero vector, so in addition to the trivial element $\phi$ in the kernel,
we have this additional element $v$, so the transformation cannot be one-one anymore.
This implies that the transform cannot be bijective, and thus the matrix associated with it must be singular.
Thus, we are left with a polynomial in $\lambda$, of degree $n$ (the dimension of the transform) at most. Upon solving this for it’s roots, which may be at most $n$ in number, we get the possible values of $\lambda$, i.e the possible eigenvalues.
Now that we have obtained the possible eigenvalues for the transform, we can substitute it into
\(M \cdot v = \lambda \cdot v\)
giving us a set of $n$ linear equations, for the $n$ unknown components of $v$, thus we have a solvable system of equations for $v$.
In this manner, we can find the corresponding eigenvectors for each eigenvalue.