# Why do determinants in this specific case can be removed?

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$- A)$

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

• Thank you. Very interesting. Where can I read more about extracting the power? It is not obvious to me that we can do it. – 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

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

• Well, not for every matrix the determinant is a real number. – Marc van Leeuwen Apr 12 '16 at 10:42
• 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. – Justin Benfield Apr 12 '16 at 11:25
• Thank you. I forgot that det() is a number actually. – Li Cooper Apr 12 '16 at 15:29