# Jordan Normal Form Question

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?

• I suppose you meant $\;A\;$ is the JCF of $\;T\;$ ...right? – DonAntonio May 30 at 14:46
• Yes, that's correct! – jessg12345 May 30 at 15:03

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.