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$A=\begin{pmatrix} -1 &-1 \\ 0& 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 &-1 \\ 0& -1 \end{pmatrix}$.

They have same trace, determinant, charecteristic polynomial and hence char poly.

But how do I know for sure they are or not similar?

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    $\begingroup$ they both have the eigenvalues $-1,1$ hence those are pairwise different hence diagonalisable hence ... $\endgroup$ – Dominic Michaelis Sep 30 '15 at 12:50
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$A=\begin{pmatrix} -1 &-1 \\ 0& 1 \end{pmatrix}$

After I do $R_1 \gets R_1 + R_2$ and $R_1\gets -R_1 $ I end up with $\begin{pmatrix} 1 &0 \\ 0& 1 \end{pmatrix}$

And for $B=\begin{pmatrix} 1 &-1 \\ 0& -1 \end{pmatrix}$. After I do $R_1 \gets R_1 + R_2$ and $R_2\gets -R_2 $ I end up with

$\begin{pmatrix} 1 &0 \\ 0& 1 \end{pmatrix}$ thus $A$ and $B$ are similar.

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  • $\begingroup$ This is not correct. Similarity means $B=P^{-1}AP$ for some invertible matrix $P$. You haven't proved the existence of $P$. What you did prove is that $A^2=B^2=I$, but that per se doesn't make $A$ and $B$ similar to each other. $\endgroup$ – user1551 Sep 30 '15 at 14:51
  • $\begingroup$ @user1551, I have shown that there is there are unitary matrices $U$ and $V$(product of elementary matrices) such that $U^{-1}AU=I=V^{-1}AV$ thus $VU^{-1}AUV^{-1}=B$, now let $P=UV^{-1}$. $\endgroup$ – R.N Sep 30 '15 at 15:21
  • $\begingroup$ Are you sure? Would you please write down explicitly $U$ and $V$? By the way, the product of a number of elementary matrices is not necessarily a unitary matrix. $\endgroup$ – user1551 Sep 30 '15 at 15:35
  • $\begingroup$ excuse me it is better that I say invrtible matrix $\endgroup$ – R.N Sep 30 '15 at 15:38
  • $\begingroup$ good point, Similarity is transitive. $\endgroup$ – R.N Sep 30 '15 at 15:56
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Take $P = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} $. Then $P^{-1}AP=B$.

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