- An nxn matrix always has n eigenvalues, counting multiplicities. While these eigenvalues may repeat, their total count, including repetitions, always sums to n.
- However, an nxn matrix does not always have n independent eigenvectors. A lack of sufficient eigenvectors can occur only when the matrix has repeated eigenvalues.
- It is important to note that repeated eigenvalues do not necessarily result in a deficit of eigenvectors. But when repeated eigenvalues are present, there is a possibility of such a deficit.
- If there is a deficit, it is equal or less than to the number of repeated eigenvalues.
Example: Consider an 12×12 matrix with eigenvalues 2, 3, 3, 7, 7, 7, 7, 6, 6, 6, 4. This matrix may have 5 eigenvectors at a minimum, as each distinct eigenvalue gets one eigenvector. It can have 12 eigenvector at a maxımum, as each eigenvalue get an eigenvector. And it can get any number of eigenvectors in between.
Example: Consider an 12×12 matrix with eigenvalues 1, 7, 4, 2, 6, 8, -1, -4, 10, 11, -10, 47. As all eigenvalues are distinct this matrix is will always have 12 eigenvectors.
A matrix can be diagonalizable only when it has $\mathrm{dim}(A)$ number of eigenvectors. In all the other cases, we will not be able to construct the matrix $P$ and our methods for diagonalization will fail.
This creates us a problem, because to solve a differential equation $y=Ax$ we need to diagonalize the nxn matrix $A$. For diagonalization, we need n eigenvectors. Hence the methods we have developed for solving matrix differential equations will not work if the matrix $A$ does not have n independent eigenvectors.