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How can I prove that $A\mathcal{R}B$ is an equivalence relation if there exists an invertible matrix $C$ such that $B = CA$?

I know there there is a reflexive, symmetric, and transitive steps.

Reflexive: $A \mathcal{R} A$ because $A = CA$, where $C = I$ this hold true?

Am I missing something important about matrices?

Thanks.

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2 Answers 2

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You have done reflexivity correctly.

For symmetry: note that $B=CA$ implies $C^{-1}B=A$.

For transitivity: If you have $B=CA$, $D=QB$, can you conclude anything about $D$? Hint: Make $D$ the subject using the previous two equations.

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  • $\begingroup$ In the transitivity argument, I believe it should be two not necessarily equal $C$'s. So $B=CA$ and $D=QB$ for invertible $C,Q$. $\endgroup$
    – String
    Apr 28, 2016 at 6:43
  • $\begingroup$ Yes, thanks. Will change it $\endgroup$
    – yoyostein
    Apr 28, 2016 at 6:44
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Reflexivity: For any matrix $A\in \mathbb{R}^{n\times n}$ we have that $A=IA$ where $I$ is the identity matrix (which obviously is invertible).

Symmetric: If $A\sim B$, then $\exists C$ such that $C$ is invertible and $A=CB$, hence $B=C^{-1}A$ where $C^{-1}$ is also invertible.

Transitivity: Suppose that $A\sim B$ and $B\sim C$. Then there exists invertible matrices $D,E$ such that $A=DB$ and $B=EC$, hence $A=D(B)=D(EC)=(DE)C$. Since $(DE)^{-1}=E^{-1}D^{-1}$, $A\sim C$. Hence this relation is transitive.

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