The eigenproblem focuses on finding the scalars, $\lambda$, and vectors $\ket v$ respectively called the eigenvalues and eigenvectors that have the solve the following problem
$$ \hat L \ket v =\lambda\ket v\\ (\hat L-\lambda \hat I)\ket v=0 $$
where $\hat L$ is a linear operator and $\ket {v}\ne0$ (i.e. we do not consider the trivial solution).
Traditionally the vectors that span an equation of this form are called the Kernel or Nullspace of the operator $\hat L-\lambda\hat I$. When an operator is invertible then its nullspace is only the trivial solution, precisely what we want to avoid (the reason for this is that for an operator to be invertible there must be a one-to-one correspondence between inputs and outputs. If more than one vector gives the output zero, if we only knew the output was zero we would not know which vector was input.)
We want our linear operator $\hat L-\lambda\hat I$ to be non-invertible which by the Invertible Matrix Theorem is equivalent to
$$ \det(\hat L-\lambda\hat I)=p(\lambda)=0 $$
This will produce a characteristic polynomial in $\lambda$. By finding the zeroes of the characteristic polynomials we find the eigenvalues of $\lambda$.
Once we have the eigenvalues, we can find the eigenvectors by solving for a given eigen value
$$ \ker(\hat L-\lambda\hat I) $$
The linearly independent vectors that span this kernel will be the eigenvectors corresponding to that eigenvalue. We always have the freedom to scale/normalize our eigenvectors.
For eigenvalues with a multiplicity of $1$ (i.e. the root $(\lambda-\lambda_1)^n$ has $n=1$), the kernel will be
$$ \ker(\hat L-\lambda_1\hat I)=\text{span}\{\ket{v_1}\} $$
For eigenvalues with a multiplicity $k>1$ we have a degeneracy and the kernel is a span of $n$ linearly independent eigenvectors such that $0\le n\le k$. We can choose eigenvectors from this span that are orthonormal through the Gram-Schmidt process. However, if $n<k$ then we will not be able to form a full spanning set of eigenvectors for our original vector space $V$. If $n=k$ for all our eigenvalues, then we can produce a linearly independent set of eigenvectors that span our vector space $V$
If an operator $\hat L$ is hermitian, then $\hat L=\hat L^\dag$, then the eigenvalues and eigenvectors will hold some special properties by the Spectral Theroem