0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ I suppose you meant $\;A\;$ is the JCF of $\;T\;$ ...right? $\endgroup$ – DonAntonio May 30 at 14:46
  • $\begingroup$ Yes, that's correct! $\endgroup$ – jessg12345 May 30 at 15:03
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.