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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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  • $\begingroup$ Which part exactly are you confused about? The equality $Z^{-1}AZ-\lambda I=Z^{-1}(A-\lambda I)Z$, or the fact that $|Z^{-1}(A-\lambda I)Z|=|Z||A-\lambda I||Z^{-1}|$? $\endgroup$ – carmichael561 Sep 4 '17 at 4:30
  • $\begingroup$ I'm confused about the first part. $\endgroup$ – Rohit Pandey Sep 4 '17 at 4:30
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    $\begingroup$ I have added (proof-explanation) tag - see the tag-info - since your questions seems to be about this specific proof (rather than asking for any proof of the claim in the title). Of course, if this is incorrect, feel free to remove this tag. $\endgroup$ – Martin Sleziak Sep 4 '17 at 5:28
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Recall that scalar matrices commute with all other matrices. Therefore $$\lambda I=\lambda IZ^{-1}Z=Z^{-1}\lambda IZ$$ It then follows that $$ Z^{-1}AZ-\lambda I=Z^{-1}AZ-Z^{-1}\lambda IZ=Z^{-1}(A-\lambda I)Z$$

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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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