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let $A\in F^{8x8}$ , $A^3=0$,$A^2\neq0$ Find its Jordan forms that possible.

solution was : characteristic polynomial $P_A(x)=x^8$and $q_8=m_A(x)=x^3,q_7=x^3,q_6=x^2,q_5=1,q_4=1,q_3=1,q_2=1,q_1=1$

so $e_1(x)=x^3,e_2(x)=x^3,e_3(x)=x^2$

I dont understand how he write $x^8$ and also writing of $q_i$. I know $m_A(x)=x^3$ and $q_1.q_2...q_8=P_A(x)$ how do we know $P_A(x)=|xI-A|$ in form of \begin{pmatrix} x & 0 & 0 & \cdots & 0 \\ a & x & 0 & \cdots & 0 \\ b & c & x & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ h & j & k & \cdots & x \end{pmatrix} thanks

similar question : $A\in F^{7x7},A^2=A $ find its Jordan which is possible

since $A^2-A=0$ $q_7=x^2-x,q_6=x^2-x,q_5=x^2-x,q_4=x,q_3=1,q_2=1,q_1=1$

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The only eigenvalue for the matrix is $0$ because $A^{3}$ is the minimal polynommial. Jordan canonical form allows you to arrange the blocks starting with the largest in the upper left, and in order of descending block size. All of the blocks have $0$'s along the diagonal and $1$'s on the diagonal above the main diagonal, unless the block size is $1$. A block $B_{n}$ with $n$ $0$'s on the diagonal and $n-1$ one's one the superdiagonal has order $n$, meaning that $B_{n}^{n}=0$ but $B_{n}^{n-1} \ne 0$. The extreme case is $B_{1}^{1}=0$.

The block sizes of all blocks must sum to $8$. The largest block size is $3$, and there must be at least one such block because $A^{3}=0$, but $A^{2}\ne 0$. That leaves $5$ for the total of the remaining block sizes. So these are the possible Jordan block sizes $$ (3,3,2),(3,3,1,1), (3,2,2,1), (3,2,1,1,1), (3,1,1,1,1,1). $$

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  • $\begingroup$ thanks. lets say $A\in F^{8x8} A^5=0,A^4\neq 0$ so $q_8=x^5$ since minimal polynommial $m_A(x)=x^5$ we can choose the sizes (5,2,1),(5,1,1,1)... right? and is $P_A(x)=x^8$ arbitrary? can it be something else? $\endgroup$ – lyme May 29 '14 at 20:24
  • $\begingroup$ If $A$ is an $8\times 8$ matrix with $A^{8}=0$ and $A^{7}\ne 0$, then there's one $8$ block, and that's the whole matrix. $\endgroup$ – DisintegratingByParts May 29 '14 at 20:28
  • $\begingroup$ my last question, what can we say about charac. polynomial of A if $A\in F^{7x7},A^2=A $ aside from $(x^2-x)|_{P_A(x)}$? $\endgroup$ – lyme May 29 '14 at 20:50
  • $\begingroup$ So you have $x(x-1)=0$. That means that the blocks with $0$ in the diagonal are simple and the blocks with $1$'s in the diagonal are also simple. So, the Jordan form is diagonal with $1$'s and $0$'s along the diagonal. There is a theorem: The minimal polynomial has distinct factors iff the matrix is diagonalizable. $\endgroup$ – DisintegratingByParts May 29 '14 at 22:10
  • $\begingroup$ Added remark: for the case mentioned $A^{2}-A=0$, it can be that $A-I=0$ or $A=0$ if $x^{2}-x$ is not the minimal polynomial. So you might have (a) all simple blocks (size 1) with $0$'s on the diagonal or (b) all simple blocks with $1$'s on the diagonal. $\endgroup$ – DisintegratingByParts May 29 '14 at 22:18
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Notice that this sequence $(\ker A^k)_k$ is strictly increasing and stationary:

$$\{0\}\subset\ker A\subset\ker A^2\subset\ker A^3=\Bbb R^8$$ moreover it's well known that the sequence $$(\dim\ker A^{k+1}-\dim\ker A^k)_k$$ is decreasing hence using the Young tableau we have these possibilities: enter image description here

and each row of Young tableau is a jordan block so for example for the last tableau we have two blocks with size $3$ and a block with size 2 hence the Jordan matrix is $(3,3,2)$.

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Note that $A^3=\mathbf0$ implies $A$ is nilpotent. It follows that the characteristic polynomial of $A$ is $$ \DeclareMathOperator{char}{char}\char_A(\lambda)=\lambda^8 $$ Furthermore, since $A^2\neq\mathbf0$, the minimal polynomial of $A$ is $$ m_A(\lambda)=\lambda^3 $$ This information tells us that the Jordan form $J$ of $A$ has

  1. Diagonal consisting only of $0$'s
  2. The size of the largest Jordan block is $\deg m_A(\lambda)=3$

A few of the possible Jordan forms are $$ \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ Can you find the rest?

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  • $\begingroup$ Thanks. appriciated. in second example that I wrote. A isnt nilpotent right? so characteristic polynomial doesnt have to be $char_A(λ)=λ^7$ ? $\endgroup$ – lyme May 29 '14 at 20:34

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