# About the rank of Jordan blocks

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!

• Hint: when $\lambda = 0$, the columns of $J_n(\lambda)$ are just standard basis vectors and one zero vector. Commented Nov 18, 2020 at 10:18
• So when $\lambda = 0$, the rank is $n-1$; and it is $n$ in all other cases? Commented Nov 18, 2020 at 10:20
• Yep!${}{}{}{}{}$ Commented Nov 18, 2020 at 10:21
• Does this have any other implications? How do I make intuitive sense of this result, in terms of Jordan strings, basis etc.? Commented Nov 18, 2020 at 10:23
• 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. Commented Nov 18, 2020 at 10:27

## 1 Answer

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