Assume a linear operator $A$ is not diagonalisable. Let $E_{\lambda i}$ be the eigenspaces corresponding to eigenvalues $\lambda_i$. I know that $E_{\lambda i}$ is an invariant subspace of $A$, and that $E_{\lambda i} \bigoplus E_{\lambda j}$ are others, but are these all the non-trivial invariant subspaces? Meaning are there invariant subspaces I could miss by going for the Eigenspaces and their combinations?
I've found a lot of similar topics to this, but I haven't found any solid proof for non-diagonalisable operators. Thanks in advance!