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Let $(A_1,B_1),(A_2,B_2) \in \Bbb R^{n\times n} \times \Bbb R^{n \times m}$. We say that $(A_1,B_1)$ and $(A_2,B_2)$ are similar written $(A_1,B_1)$ ~ $(A_2,B_2)$, if there exists $S \in$ GL (n, $\Bbb R$) such that $A_2=S^{-1}A_1S$ and $B_2 = S^{-1}B_1$. How can I show that ~ is an equivalence relation?

I know that I have to show the three properties of equivalence relation:

  1. symmetric
  2. reflexive
  3. transitive

But I do not know what to do to conclude to these.

Thank you in advance.

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1 Answer 1

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Symmetry:

If $(A_1,B_1)\sim (A_2,B_2)$ this means $A_2=S^{-1}A_1S$ and $B_2 = S^{-1}B_1S$. Now you can choose $S'=S^{-1}$ and you get $(A_2,B_2)\sim (A_1,B_1)$

Reflexivity

Choose $S=I$ the identity than you get that $A_1=S^{-1}A_1 S$ and $B_1=S^{-1}B_1 S$. So you obtain $(A_1,B_1)\sim (A_1,B_1)$

Transitive

This means: Assume $(A_1,B_1)\sim (A_2,B_2)$ and $(A_2,B_2)\sim (A_3,B_3)$ then it holds that $(A_1,B_1)\sim (A_3,B_3)$.

If the first is done with $S'$ and the second with $S''$ then the third may be done with $S'S''$

Because all these conditions are fullfilled you know this is a equivalence relation.

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  • $\begingroup$ For the $B$ matrix it isn't a similarity transformation, i.e. $B_2 = S^{-1} B_1$. It comes from the coordinate change of states in a LTI dynamical system. I just wanted to point out the mistake, your arguments still hold in this case too, anyway. $\endgroup$
    – obareey
    Commented Nov 4, 2015 at 9:53
  • $\begingroup$ Is there no $S$ missing after $B_1$. Thought this is a typo. $\endgroup$
    – Matthias
    Commented Nov 4, 2015 at 10:02
  • $\begingroup$ I don't think it's a typo, as I explained the reason. Also $B$ has dimensions $n \times m$, so there shouldn't be an $S$ after $B_1$. $\endgroup$
    – obareey
    Commented Nov 4, 2015 at 10:13

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