Let $A$ be a $n\times n$ matrix. If $\lambda$ is an eigenvalue of A with multiplicity $m>1$. Then there could be two cases:
- There are $m$ linearly independent eigenvectors of A corresponding to $\lambda$.In this case, the geometric multiplicity is the same as the algebraic multiplicity.
- The geometric multiplicity is less than algebraic multiplicity.
This information is quite useful while solving a system of linear ODEs. If we have the situation in case 2, we resort to finding generalized eigenvectors.
For example, suppose A is a $3\times 3$ matrix with repeated eigenvalue $\lambda = 1$ and we found only a single eigenvector $v_1$ by solving $(A-\lambda I)v_1 = 0$. A nontrivial solution is always guaranteed for a homogeneous case $Bx = 0$, where $B$ is singular.
Now, we find the generalized eigenvalue of rank 2 by solving the equation $$(A-\lambda I)v_2 = v_1$$.
See that $(A-\lambda I)$ is singular and it's a non-homogeneous case. Why is it so that the system is not inconsistent? In fact $$(A - \lambda I)^mv = 0$$ for a nonzero $v$ makes it possible to repeat the process of finding generalized eigenvectors up to $m-1$ steps. What makes $(A - \lambda I)^mv = 0$ possible.
Can anyone clear it in simple language? Thanks in advance.