Proof of Theorem ETM

Suppose $A$ has size $n$. Then \begin{align*} \charpoly{A}{x} &=\detname{A-xI_n}&&\knowl{./knowls/definition.CP.knowl}{\text{Definition CP}}\\ &=\detname{\transpose{\left(A-xI_n\right)}}&&\knowl{./knowls/theorem.DT.knowl}{\text{Theorem DT}}\\ &=\detname{\transpose{A}-\transpose{\left(xI_n\right)}}&&\knowl{./knowls/theorem.TMA.knowl}{\text{Theorem TMA}}\\ &=\detname{\transpose{A}-x\transpose{I_n}}&&\knowl{./knowls/theorem.TMSM.knowl}{\text{Theorem TMSM}}\\ &=\detname{\transpose{A}-xI_n}&&\knowl{./knowls/definition.IM.knowl}{\text{Definition IM}}\\ &=\charpoly{\transpose{A}}{x}&&\knowl{./knowls/definition.CP.knowl}{\text{Definition CP}}\\ \end{align*}

So $A$ and $\transpose{A}$ have the same characteristic polynomial, and by Theorem EMRCP, their eigenvalues are identical and have equal algebraic multiplicities. Notice that what we have proved here is a bit stronger than the stated conclusion in the theorem.