Proof of Theorem SER

To see that $A$ is similar to $A$, we need only demonstrate a nonsingular matrix that effects a similarity transformation of $A$ to $A$. $I_n$ is nonsingular (since it row-reduces to the identity matrix, Theorem NMRRI), and \begin{equation*} \similar{A}{I_n}=I_nAI_n=A \end{equation*}

If we assume that $A$ is similar to $B$, then we know there is a nonsingular matrix $S$ so that $A=\similar{B}{S}$ by Definition SIM. By Theorem MIMI, $\inverse{S}$ is invertible, and by Theorem NI is therefore nonsingular. So \begin{align*} \similar{A}{(\inverse{S})}&=SA\inverse{S}&&\knowl{./knowls/theorem.MIMI.knowl}{\text{Theorem MIMI}}\\ &=S\similar{B}{S}\inverse{S}&&\knowl{./knowls/definition.SIM.knowl}{\text{Definition SIM}}\\ &=\left(S\inverse{S}\right)B\left(S\inverse{S}\right)&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}}\\ &=I_nBI_n&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=B&&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}} \end{align*} and we see that $B$ is similar to $A$.

Assume that $A$ is similar to $B$, and $B$ is similar to $C$. This gives us the existence of two nonsingular matrices, $S$ and $R$, such that $A=\similar{B}{S}$ and $B=\similar{C}{R}$, by Definition SIM. (Notice how we have to assume $S\neq R$, as will usually be the case.) Since $S$ and $R$ are invertible, so too $RS$ is invertible by Theorem SS and then nonsingular by Theorem NI. Now \begin{align*} \similar{C}{(RS)}&=\similar{\similar{C}{R}}{S}&&\knowl{./knowls/theorem.SS.knowl}{\text{Theorem SS}}\\ &=\similar{\left(\similar{C}{R}\right)}{S}&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}}\\ &=\similar{B}{S}&&\knowl{./knowls/definition.SIM.knowl}{\text{Definition SIM}}\\ &=A \end{align*} so $A$ is similar to $C$ via the nonsingular matrix $RS$.