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Using the Jordan canonical form, show given any square matrix A and e>0, that there exists a matrix Q that

$Q^{-1}AQ=\pmatrix{J^{1}_e&0&...&0\\0&J^{2}_e&...&0\\...&...&...&...\\0&...&...&J^{q}_e}$ where each block $J^{i}_e$ = mI + eN.

I have no idea about how to prove that... I wanted to get some ideas from my previous question but failed. previous one

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Hint: if $J$ is an $n$-by-$n$ Jordan block with eigenvalue $\lambda$ and $D = \text{diag}(1,\epsilon,\dots,\epsilon^{n-1})$, compute $D^{-1} J D$.

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  • $\begingroup$ Sorry.. I just can't understand what you mean... $\endgroup$
    – Sam
    Commented Mar 5, 2018 at 16:01

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