Given A $\in \Bbb F^{14 \times 14}$ such that $rank (A^4)=0$ $rank (A^3)=1$ $rank (A^2)=4$ $rank (A)=8$ find the matrix Jordan normal form
The answer in the textbook is $J_A=diag(J(0,4),J(0,3),J(0,3),J(0,2),J(0,1),J(0,1))$
I am trying to understand how they got to that answer.. I assumed it has something to do with the following formula for finding the number of jordan chains $s_i=\dim(ker(T- \lambda I)^i)-\dim(ker(T- \lambda I)^{i-1})$
But if I do that I would get $s_4=14-13=1$ so there is only 1 jordan block of size 4 $s_3=13-10=3$ 3 blocks of size 3? but in the book it says 2 blocks according to the answer $s_2=10-6=4$ and $s_1=6-0=6$
I think that I do not understand quite well what the formula actually means and also $rank(J(0,k)^i)=k-i$
Would really appreciate an explanation to $rank(J(0,k)^i)=k-i$ and $s_i=\dim(ker(T- \lambda I)^i)-\dim(ker(T- \lambda I)^{i-1})$
what do the number I get mean ? how would I use it ?
I thought it meant how many blocks there are for $s_i=\dim(ker(T- \lambda I)^i)-\dim(ker(T- \lambda I)^{i-1})$
but for example $s_3=3 $ but in the answer there are only $2$ blocks
thanks for any tips and help !
Your formula may be wrong.
If we let $\lambda_1,\lambda_2,\cdots,\lambda_s$ be all distinct eigenvalues of A, then number of jordan block of size $t$ with $\lambda_j$ as the diagonal element is $$N_j\left( t \right) =\mathrm{rank}\left( A-\lambda _tI \right) ^{t+1}+\mathrm{rank}\left( A-\lambda _tI \right) ^{t-1}-2\mathrm{rank}\left( A-\lambda _tI \right) ^t.$$
Back to this question, $0$ is the unique eigenvalue of $A$. So $N_0\left( 4 \right)=0+1-2\times0=1$,$N_0\left( 3 \right)=0+4-2=2$, $N_0\left( 2 \right)=1+8-8=1$, $N_0\left( 1 \right)=14+4-16=2$.
So $$J_A=diag(J(0,4),J(0,3),J(0,3),J(0,2),J(0,1),J(0,1))$$
Some explanations:
Just use $\mathrm{rank}\left( J\left( 0,k \right) ^i \right) =\begin{cases} k-i,\,\, i<k\\ 0, ~~~i\geqslant k\\ \end{cases}$. Then you can get the formula.
For example, If $B$ is a matirx and $B^l=0$. We denote $N(t)=N_0(t)$ $(N(l+1)=N(l+2)=0).$
Consider $J_B^{t-1}$, the jordan block of size less than $t-1$ is $0$.
Then $\mathrm{rank}(J_B^{t-1})=N(t)[t-(t-1)]+N(t+1)[t+1-(t-1)]+\cdots+N(l)[l-(t-1)]+N(l+1)[l+1-(t-1)]+N(l+2)[l+2-(t-1)]\\=N(t)+2N(t)+\cdots+N(l)(l-t+1)+N(l+1)(l-t+2)+N(l+2)(l-t+3)(*).$
Substitute $t+1$ for $t$ in $(*)$ , we get $\mathrm{rank}(J_B^{t})=N(t+1)+2N(t+2)+\cdots+N(l)(l-t)+N(l+1)(l-t+1)+N(l+2)(l-t+2)(**).$
$(*)-(**)$, we can get $\mathrm{rank}(J_B^{t-1})-\mathrm{rank}(J_B^{t})=N(t)+N(t+1)+\cdots+N(l)(***)$.
Substitute $t+1$ for $t$ in $(***)$, we can get $\mathrm{rank}(J_B^{t})-\mathrm{rank}(J_B^{t+1})=N(t+1)+N(t+2)+\cdots+N(l)(****)$.
$(***)-(****)$ is $N(t)=\mathrm{rank}(J_B^{t+1})+\mathrm{rank}(J_B^{t-1})-2\mathrm{rank}(J_B^{t})$.
For any matrix $A$, it is the same.