Is it always the case that for non-defective matrices the geometric multiplicities (dimensions of the eigenspaces) of the eigenvalues will equal the algebraic multiplicities ? What theorems are used in proving this statement ?
Take the $n \times n$ matrix $A$ with $\lambda_{1} \dots \lambda_{k}$ distinct eigenvalues but $k < n$. For each eigenvalue, if the dimension of the eigenspace $\varepsilon_{A}(\lambda_{k})$ is $d_{k}$ and $\sum\limits_{k} d_{k} = n$, then the matrix is non-defective and it is possible to build a set of linearly independent eigenvectors that spans $\mathbb{R}^{n}$ (i.e a basis) to diagonalize A.
Collecting this basis in the matrix $P$, we can diagonalize the matrix via the similarity transformation $D = P^{-1}AP$.