Good evening everyone,

Why do determinants on both sides in the case below can be removed? And what do we actually do in order to achieve it? Thank you.

$\det(S^{-1})\det(\lambda$I$ - A)\det(S) = \det(\lambda$I$ - A)$

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    $\begingroup$ It depends on your country. In mine you could get a 5 year sentence! $\endgroup$ – YoTengoUnLCD Apr 12 '16 at 3:54
  • $\begingroup$ @YoTengoUnLCD Which crime can you be charged with for illegally removing determinants? Destruction of public properties? $\endgroup$ – Batominovski Apr 12 '16 at 3:57

For any invertible matrix $S$, $\det(S^{-1}) = \det(S)^{-1}$, so the $\det(S^{-1})$ and $\det(S)$ cancel.

  • $\begingroup$ Thank you. Very interesting. Where can I read more about extracting the power? It is not obvious to me that we can do it. $\endgroup$ – Li Cooper Apr 12 '16 at 15:29

Note that for any square matrix $M$, $\det(M)$ is always a real number, hence

$\det(S^{-1})\det(\lambda I-A)\det(S)=\det(S^{-1})\det(S)\det(\lambda I-A)$

by commutativity of real number multiplication. Now since $\det(S^{-1})=(\det(S))^{-1}$, or in otherwords, $\det(S^{-1})=\frac{1}{\det(S)}$, it follows that

$\det(S^{-1})\det(S)\det(\lambda I-A)=\det(\lambda I-A)$.

  • $\begingroup$ Well, not for every matrix the determinant is a real number. $\endgroup$ – Marc van Leeuwen Apr 12 '16 at 10:42
  • $\begingroup$ I took it as an assumption that the entries of the matrices were real numbers. Indeed matrices whose entries come from another field, such as $\mathbb{C}$ can be non-real, but they will come from whichever field the entries of the matrices came from. $\endgroup$ – Justin Benfield Apr 12 '16 at 11:25
  • $\begingroup$ Thank you. I forgot that det() is a number actually. $\endgroup$ – Li Cooper Apr 12 '16 at 15:29

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