What can be said about the rank of Jordan blocks of size $n$? When is rank $J_n(\lambda) = n$?

From my observation, it seems that whenever $\lambda\neq 0$, the corresponding Jordan block has full rank (i.e. $n$). This seems clear from the fact that $J_n(\lambda)$ is upper triangular, and if we select any two columns $c_1,c_2$ and put $ac_1 + bc_2 = 0$, then we must have $a=b=0$. However this fails when $\lambda = 0$, and the matrix becomes strictly upper triangular.

I would appreciate any hints on how to be able to make generalized comments about the rank of Jordan blocks. Thank you!

  • $\begingroup$ Hint: when $\lambda = 0$, the columns of $J_n(\lambda)$ are just standard basis vectors and one zero vector. $\endgroup$
    – user847970
    Commented Nov 18, 2020 at 10:18
  • $\begingroup$ So when $\lambda = 0$, the rank is $n-1$; and it is $n$ in all other cases? $\endgroup$ Commented Nov 18, 2020 at 10:20
  • $\begingroup$ Yep!${}{}{}{}{}$ $\endgroup$
    – user847970
    Commented Nov 18, 2020 at 10:21
  • $\begingroup$ Does this have any other implications? How do I make intuitive sense of this result, in terms of Jordan strings, basis etc.? $\endgroup$ Commented Nov 18, 2020 at 10:23
  • 1
    $\begingroup$ It has no implications I know of that aren't easier to see from other perspectives, but it helps solidify two things: first, the non-zero eigenvalues and their generalised eigenspaces do not diminish the rank of the matrix, and second, the chains of generalised eigenvectors corresponding to $0$ shorten by one when multiplying by the matrix. This isn't a particularly deep result; I think it's more of a thought exercise to help you wrap your head around Jordan blocks and how they work. $\endgroup$
    – user847970
    Commented Nov 18, 2020 at 10:27

1 Answer 1


I am adding this answer just for the sake of completeness, it is definitely trivial and unnecessary:

$$\text{rank }J_n(\lambda)= n-1 \text{ if }\lambda=0$$ and $$\text{rank }J_n(\lambda)=n \text{ if }\lambda\neq 0$$


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