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I have the two matrices:

$\begin{pmatrix}1&-4&-2\\ 0&1&0\\ 0&4&3\end{pmatrix}$ and $\begin{pmatrix}3&0&0\\ \:0&1&1\\ \:0&0&1\end{pmatrix}$

I know they have the same trace and determinant but I know that isn't enough to prove they are similar... what are the next steps I should take? Is proving they have the same eigenvalues enough to show they are similar?

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    $\begingroup$ The first matrix is already in Jordan normal form... $\endgroup$ – user251257 Feb 28 '18 at 20:02
  • $\begingroup$ Do you have studied matrix reduction? $\endgroup$ – user371663 Feb 28 '18 at 20:03
  • $\begingroup$ For characteristic $2$ the first matrix is diagonal, but the second is not diagonalizable. Hence they cannot be similar in this case. $\endgroup$ – Dietrich Burde Feb 28 '18 at 20:10
  • $\begingroup$ And spoiler: they aren't similar. Just compute the Jordan normal form. $\endgroup$ – user251257 Feb 28 '18 at 20:10
  • $\begingroup$ @DietrichBurde nope. The first matrix - $I$ has kernel dimension 2, unlike the second one. $\endgroup$ – user251257 Feb 28 '18 at 20:15
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Hint: The definition of similarity between matrices is the following:

Two square matrices of the same dimensions A and B are said to be similar if there is a matrix P such that $$B = P^{-1}AP$$

Try finding a matrix $P$ for your exercise.

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Hint: A matrix $A$ and $B$ are said to be similar if $$B=P^{-1}AP$$ for some invertible matrix $P$. This link: How do I tell if matrices are similar? outlines the entire process.

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The first matrix is diagonalizable, namely $P^{-1}AP={\rm diag}(1,1,3)$ with $$ P=\begin{pmatrix} -3 & 2 & -2\cr -2 & 1 & 0 \cr 4 & -2 & 2 \end{pmatrix}. $$ The second matrix is not diagonalizable, so they are in fact not similar.

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