Matrix similarity is an equivalence relation

Question

Recall that for $ A,B \in \mathsf{M}_n$, we say that $A$ is similar to $B$, denoted $A \sim B$, if there is an invertible matrix $S \in \mathsf{M}_n$ such that $A = S B S^{-1}$. Prove that similarity is an equivalence relation on $\mathsf{M}_n$.

I have no idea on what to do here, what i have tried is to say that $A = S A S^{-1} = A I_n = B = BI_n = S B S^{-1} $ which i have a very strong feeling is completely wrong. What should i do? how do i do this proof?

if yall can come up with a better title feel free to replace or change mine.

Answer 1

I'm not sure how you get any of the equalities in your work, except for $B=BI_n$. Since $S$ is just some invertible matrix, how do you get $A=SAS^{-1}=AI_n$ and $BI_n=SBS^{-1}$? I suspect you are commuting $SAS^{-1}$ into $ASS^{-1}$ and similarly $SBS^{-1}$ into $BSS^{-1}$, but remember that matrix multiplication is not commutative. For example, if $$ S:=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \quad\text{and}\quad B:=\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, $$ then $$ SBS^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1/2 & 1 \end{pmatrix} $$ is not equal to $B$. Also, when you write $AI_n=B$, you seem to be assuming that $A$ is equal to $B$. But this isn't true, even for similar matrices. For instance, if $$A:=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \quad\text{and}\quad B:=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \quad\text{and}\quad S:=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} ,$$ then $A\ne B$ are similar because $S$ is invertible and $$ SBS^{-1} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} =\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = A. $$

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Now, onto the question. We need to prove that $\sim$ is reflexive ($A\sim A$ for all $A\in\mathsf{M}_n$), symmetric ($A\sim B$ implies $B\sim A$), and transitive ($A\sim B$ and $B\sim C$ implies $A\sim C$).

I'll let you do reflexivity.

For symmetry, suppose $A\sim B$. Then there exists an invertible matrix $S$ such that $A=SBS^{-1}$. Hence $S^{-1}$ is invertible and $B=S^{-1}A(S^{-1})^{-1}$, so $B\sim A$.

For transitivity, assume $A\sim B$ and $B\sim C$. Then there are invertible matrices $S$ and $T$ such that $A=SBS^{-1}$ and $B=TCT^{-1}$. Then $ST$ is invertible and $$ A = SBS^{-1} = S(TCT^{-1})S^{-1} = (ST)C(ST)^{-1}, $$ so $A\sim C$.