Given a finite dimensional vector space $V$ over $\mathbb{C}$ and a linear transformation $L:V \to V $, we know that eigenvalues exist, since the characteristic equation will be a polynomial with complex coefficients, which we know has roots. However, when do generalized eigenvectors exist?
For example, if $A$ is a real matrix, and we find that it doesn't have linearly independent eigenvectors, does it necessarily have generalized eigenvectors? It seems like the purpose of generalized eigenvectors is to find a way around things like multiplicities and linearly dependencies, but how do we know they exist?
Note that an eigenvector is in particular a generalized eigenvector. So, existence of eigenvalues implies existence of eigenvector, which in turn implies existence of generalized eigenvector.
Added Remarks about generalised eigenvectors
And, as for the purpose of generalised eigenvectors, here's a brief overview: the purpose is to get as close to diagonalizability as possible. If $T: V \to V$ has the distinct eigenvalues $\lambda_1, \dots, \lambda_r$, and is diagonalizable, our vector space has a nice direct sum decomposition: \begin{equation} V = \bigoplus_{i=1}^r E_{\lambda_i}, \end{equation} where $E_{\lambda_i} = \ker (T- \lambda_i I)$ is the eigenspace. But, when $T$ is not diagonalizable, what this means is atleast one of the eigenspaces is not "big enough" or said differently, there aren't enough eigenvectors to span $V$. So, naturally, we might be led to seek a generalization of $E_{\lambda_i}$, to a "bigger space", $G_{\lambda_i}$ which contains $E_{\lambda_i}$, and such that we have a direct sum decomposition of the vector space: \begin{equation} V = \bigoplus_{i=1}^r G_{\lambda_i}. \end{equation}
It turns out that $G_{\lambda_i} = \bigcup \limits_{q=1}^{\infty} \ker(T-\lambda_i I)^q$, satisfies this property, and this leads to the theory of Jordan Canonical Forms for linear maps whose characteristic polynomials split.