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I'm trying to find Jordan Normal Form of a linear transformation $F: V \to V$ with characteristic polynomial $$P_{F}(t) =(t+1)^3 (t-1)^3$$ and minimal polynomial $$M_{F}(t) =(t+1)^2 (t-1)^2$$

It has eigenvalues $t = -1$ and $t=1$ with algebraic multiplicities of $3$, so the JF has two $3$ x $3$ Blocks with $1$ and $-1$ in the diagonal respectively.

Since each Eigenvalue has multiplicity of $2$ in minimal Polynomial, then the size of the biggest Jordan Block to each eigenvalue is $2$ (is this correct?)

So $J_{F}$ is \begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 \end{bmatrix}

Just want to vertify my answer. Thanks.

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  • $\begingroup$ That's perfect. $\endgroup$ – Arnaud Mortier Jun 12 '18 at 3:07
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    $\begingroup$ Correct. For your question, keep in mind that, for a given characteristic root $\lambda_0$, the size of the biggest Jordan block corresponding to $\lambda_0$ is the exponent of the minimal polynomial factor $(t - \lambda_0)^{m_0}$. $\endgroup$ – Zhanxiong Jun 12 '18 at 3:11

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