A= $$ \left[ \begin{array}{ccc} 2&1&4\\ 0&1&2\\ 0&0&1 \end{array} \right] $$ and B=$$ \left[ \begin{array}{ccc} 2&0&3\sqrt2\\ 1&1&\sqrt2\\ 0&0&1 \end{array} \right] $$ Is there any method to find if they are matrix equivalent? Or we need to find a matrix S by trial and error such that $$S^*AS=B$$

  • $\begingroup$ First off, apply a permutation similarity to bring $B$ to the form $$ \pmatrix{1&1&\sqrt{2}\\0&2& 3 \sqrt{2}\\ 0&0&1} $$ $\endgroup$ – Ben Grossmann Feb 24 '17 at 6:13
  • $\begingroup$ It seems like the answer will be yes, which I find surprising $\endgroup$ – Ben Grossmann Feb 24 '17 at 6:19
  • $\begingroup$ A common trick is to compare $tr(w(A,A^*))$ and $tr(w(B,B^*))$ where $w$ denotes a word on two letters. So far, $tr(A^nA^{*n}) = tr(B^nB^{*n})$ up to $n=3$. $\endgroup$ – Ben Grossmann Feb 24 '17 at 6:21

Guess: try calculating $S^*BS$ with $$ S = \frac 1{\sqrt{2}}\pmatrix{1&1&0\\1&-1&0\\0&0&\sqrt{2}} $$ The idea is that I begin by upper triangularizing $B$ to match the form of $A$, with $2$ as the first eigenvector.

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