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let characteristic polynomial $P_A(x)=(x+2)^4(x-3)^2$ and

minimal polynomial $m_A=(x+2)^3(x-3)$ find the jordan form that possible.

we know $q_6=\frac{f_6}{f_5}$ ($f_i$ is gcd{det of i x i submatrices which isnt equal to 0})

$q_6=\frac{f_6}{f_5}=\frac{(x+2)^4(x-3)^2}{f_5}=(x+2)^3(x-3)^2$ so $f_5=(x+2)(x-3)$

I know $q_5.q_4.q_3.q_2.q_1=(x+2)(x-3)$

how do we find the rest?$ (q_5,q_4,q_3...) $

can we choose them arbitrary like $q_5=x+2,q_4=x-3,q_3=q_2=q_1=1$ or $q_5=(x-3)(x+2),q_4=q_3=q_2=q_1=1$ ?

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  • $\begingroup$ Can you elaborate a bit better how we get these polynomials $f_5,f_6,q_6$ and what are they doing to the original matrix $A$? $\endgroup$ – Berci May 29 '14 at 21:29
  • $\begingroup$ @Berci The OP is apparently using the not so known algorithm to find the jordan normal form of a matrix by resourcing to its smith normal form. See an example here. $\endgroup$ – Git Gud May 29 '14 at 21:31
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We are given that the characteristic and minimal polynomial of a matrix $A$ are \begin{align*} p_A(x) &= (x+2)^4(x-3)^2 & m_A(x) &= (x+2)^3(x-3) \end{align*} This tells us that $A$ is a $6\times 6$ matrix whose eigenvalues are $\lambda_1=-2$ and $\lambda_2=3$. The Jordan form of $A$ is thus of the form $$ \begin{bmatrix} -2 & * & 0 & 0 & 0 & 0 \\ 0 & -2 & * & 0 & 0 & 0 \\ 0 & 0 & -2 & * & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & * \\ 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} $$ The minimal polynomial of $A$ tells us that the largest Jordan block associated to $\lambda_1=-2$ has size $3$ and the largest Jordan block associated to $\lambda_2=3$ has size $1$. This gives one possible Jordan form (up to re-ordering): $$ \begin{bmatrix} -2 & 1 & 0 & 0 & 0 & 0 \\ 0 & -2 & 1 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{bmatrix} $$

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  • $\begingroup$ nicely written, I like your use of $*$ here. $\endgroup$ – James S. Cook May 29 '14 at 21:32

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