It is posited (in chapter $11$ of the book Numerical Recipes in C) that doing a similarity transform of a matrix $A$ preserves its eigenvalues. This is how a similarity transform is defined:
$$A \mapsto Z^{-1}AZ$$ for some transformation matrix $Z$ whose determinant is one. The proof given uses the characteristic equation.
$$|Z^{-1} A Z - \lambda I| = |Z^{-1}(A-\lambda I)Z| $$ I don't understand how this comes about. Once we accept this, the result follows quite easily:
$$= |Z||A-\lambda I||Z^{-1}| = |A-\lambda I|.$$
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$\begingroup$Recall that scalar matrices commute with all other matrices. Therefore $$\lambda I=\lambda IZ^{-1}Z=Z^{-1}\lambda IZ$$ It then follows that $$ Z^{-1}AZ-\lambda I=Z^{-1}AZ-Z^{-1}\lambda IZ=Z^{-1}(A-\lambda I)Z$$
$\endgroup$ $\begingroup$I'll write $\det(A)$ to mean the determinant $|A|$ of $A$. Then
\begin{align*} \det(Z^{-1}AZ-\lambda I) &= \det(Z^{-1}AZ-\lambda Z^{-1}Z) \\ &= \det(Z^{-1}AZ- Z^{-1}\lambda I Z) \\ &= \det(Z^{-1})\det(A- \lambda I )\det(Z) \\ &= \det(Z)\det(Z^{-1})\det(A- \lambda I ) \\ &= \det(A- \lambda I ). \\ \end{align*}
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