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Let $p_a=(x-2)^2(x-7)^4x$ be the characteristic polynomial of the matrix $A$ and $(x-2)^2(x-7)x$ the minimal polynomial. Determine the matrix $A$.

My work: I know the matrix has to be $7x7$ and in its diagonal it must have two $2$, four $7$ and one $0$, so:

\begin{bmatrix}{} 2& & & & & & \\ & 2& & & & &\\ & & 7 & & & &\\ & & & 7 & & &\\ & & & & 7& & \\ & & & & & 7 &\\ & & & & & & 0\\ \end{bmatrix}

I don't know how to follow, what information gives me the minimal polynomial?

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  • $\begingroup$ Do you mean determine the Jordan Form of $A$? This information does not uniquely determine $A$. $\endgroup$ – Ken Duna Apr 5 '17 at 20:47
  • $\begingroup$ @KenDuna The problem statement does not say if it is the Jordan Form of A, it just says the matrix A. However I think the matrix A is itself a Jordan Form. $\endgroup$ – J doeoeo Apr 5 '17 at 20:50
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The minimal polynomial in this case gives you the information about the relevant Jordan blocks. Since it has $(x-2)^2$ as a factor, you must have one $2 \times 2$ Jordan block associated to the eigenvalue $2$ (and not two $1 \times 1$ Jordan blocks). To see why, note that the minimal polynomial of

$$ \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} $$

is $(x - 2)$ while the minimal polynomial of

$$ \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} $$

is $(x - 2)^2$.

Similarly, since the minimal polynomial has $(x-7)$ as a factor, al the Jordan blocks associated to the eigenvalue $7$ must be $1 \times 1$. Hence, $A$ is similar to the matrix

$$ \begin{pmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 7 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 7 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}. $$

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  • $\begingroup$ Thank you for your answer, I think I understood it. Now, this means that if the minimal polynomial was $(x-2)^2(x-7)^2x$, the matrix could be with only one $1$ below the first $7$ or with one $1$ below the first $7$ and another $1$ below the third $7$? $\endgroup$ – J doeoeo Apr 5 '17 at 21:21
  • $\begingroup$ It would be impossible to decide - both options are legit. In general, the characteristic polynomial and the minimal polynomial are not enough to determine the Jordan form. $\endgroup$ – levap Apr 5 '17 at 23:12

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