Consider a complex matrix A of size $n×n$ with one of its eigenvectors, v, and the eigenvalue associated to v, denoted by $\lambda$. How to determine the length ($m$) of the Jordan chain {v$_1$,v$_2$,...,v$_m$} associated to v? I suppose if v$_m$ is the "last" generalized eigenvector in the chain, then the system of linear equations $$\left( \mathbf{A}-\lambda \mathbf{I} \right){{\mathbf{v}}_{m+1}}={{\mathbf{v}}_{m}}$$ is not solvable to v$_{m+1}$, where I is the identity matrix of size $n×n$. Is it true or another protocol is necessary to obtain $m$?
2026-04-01 19:01:38.1775070098
How to determine the length of the Jordan chain associated to an eigenvector?
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If I understand correctly, the following books say the same:
1) on the page 29: https://books.google.hu/books?id=EF3ACQAAQBAJ&pg=PA29&lpg=PA29&dq=jordan+chain+not+solvable+next+vector&source=bl&ots=hbycJE2BbQ&sig=U73q864Al-bqoBYl7l5UHkhtVt0&hl=hu&sa=X&ved=0ahUKEwi9rqvR657VAhVpJMAKHWu1ACkQ6AEIQDAE#v=onepage&q=unsolvable&f=false
2) on the page 6-7: https://books.google.hu/books?id=H40qXJ7wY1IC&pg=PA7&lpg=PA7&dq=jordan+chain+unsolvable+next+vector&source=bl&ots=4U8b_pyLN1&sig=eWa-b9rynRKBK6rXzvsVA-j5aIU&hl=hu&sa=X&ved=0ahUKEwiympOa7J7VAhXDIMAKHWQYAtQQ6AEISzAF#v=onepage&q=jordan%20chain%20unsolvable%20next%20vector&f=false
3) on the page 20: https://books.google.hu/books?id=_HetDgAAQBAJ&pg=PA30&lpg=PA30&dq=jordan+chain+unsolvable+next+vector&source=bl&ots=7X3FIDLJvh&sig=5PMIhHupqGZQtYahavF9YlglGYQ&hl=hu&sa=X&ved=0ahUKEwiympOa7J7VAhXDIMAKHWQYAtQQ6AEIUDAG#v=onepage&q=jordan%20chain%20unsolvable%20next%20vector&f=false. Is it true or I am wrong?