Assume we have a matrix $A$ with eigenvalues $\lambda_i$ and each eigenvalue has the geoemtrical multiplicity $g_i$ and the algebraic multiplicity $a_i$.
Now assume, that for a specific $i$ we have $g_i=n\cdot a_i$ with $n\in\mathbb N$
E.g. $g_1=2, a_1=4$
Notation: $J_{size,\lambda_i}$
So, for the given example, we would need $2$ blocks, since the geometrical multiplicity is $2$.
The possibilities are:
- $J_{2,\lambda_i}\boxplus J_{2,\lambda_i}$
- $J_{1,\lambda_i}\boxplus J_{2,\lambda_i}$
- $J_{2,\lambda_i}\boxplus J_{1,\lambda_i}$
I'm wondering, how I decide if I should take $(1)$, $(2)$/$(3)$. [I know, that $(2)$ and $(3)$ are equivalent]. But I'm not sure if I can always choose between $(1)$ and $(2)$ or if there are criterias which limit me in my choice.
I think you have written $J_{2,\lambda_i}\boxplus J_{1,\lambda_i}$ and $J_{1,\lambda_i}\boxplus J_{2,\lambda_i}$ instead of $J_{3,\lambda_i}\boxplus J_{1,\lambda_i}$ and $J_{1,\lambda_i}\boxplus J_{3,\lambda_i}$, rispectively.
By the way, the trick consists in calculating $(A-\lambda_i Id)^2$ and looking at its rank, which I'll write as $rank(A-\lambda_i Id)^2:=r(A-\lambda_i Id)^2$.
If $r(A-\lambda_i Id)^2 = r(A-\lambda_i Id)-2$, then you have case $(1)$, otherwise $r(A-\lambda_i Id)^2 = r(A-\lambda_i Id)-1$ and you have case $(2)$ or $(3)$.
Why? Because the rank of the squared matrix decreases by $1$ for every block with size $\ge2$.