I'm trying to properly get to know the Jordan normal form Theorem, and am confused as to why this proposition holds. I have read that if A is a matrix in Jordan normal form and $T:V\rightarrow V$ then $$\dim\ker(T-\lambda I)^r-\dim\ker(T-\lambda I)^{r-1}$$ is the number of Jordan blocks $J_m(\lambda)$ of A with $m\geq r$.

I am not sure why this holds, can anyone explain this for me?

  • 1
    $\begingroup$ I suppose you meant $\;A\;$ is the JCF of $\;T\;$ ...right? $\endgroup$
    – DonAntonio
    Commented May 30, 2019 at 14:46
  • $\begingroup$ Yes, that's correct! $\endgroup$
    – xyz12345
    Commented May 30, 2019 at 15:03

1 Answer 1


Suppose $J$ is a Jordan block of size $m$, then $\dim \ker(J-\lambda I)^r = \min (m,r)$.

Hence, if we set $r=1,2,...$, we get the dimensions $1,2,...,m-1,m,m,...$.

If we look at $\min(m,r-1)$ we get the numbers $0,1,...,m-2,m-1,m,...$

Hence $\min (m,r) - \min (m,r-1) $ gives the numbers $1,1,...,1,1,0,...$ where the transition to $0$ occurs at $r=m+1$.

Hence $\min (m,r) - \min (m,r-1) =1$ iff $r \le m$.

Summing over the blocks corresponding to $\lambda$ gives the desired result.


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