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Using concepts of diagonalization where $X = P Y P^{-1}$, I have to show that $\det(X) = \det(Y)$.
I do understand that $PP^{-1} = I$ and since $\det(I) = 1$.
So how would I correctly explain that both these statements help prove the given question?

Widawensen
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KleinESK
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    If you have been given that $\det(AB)=\det(A)\det(B)$, then this follows almost immediately. You should ensure that you have a proof for this other theorem though that doesn't rely on the theorem you want to prove in order to avoid circular logic. – JMoravitz Mar 26 '18 at 08:17
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    You should always include some context when you ask a question. JMoravitz's comment and OCH's answer are both an excellent demonstration of why! Noone knows what you know - so tell us where you found this question. for example, maybe it was it an exercise from your first linear algebra lecture, or perhaps from your last?... – user1729 Mar 26 '18 at 08:21
  • We can use $XP=PY$. – Widawensen Mar 26 '18 at 13:53

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We have the rule $\det(AB) = \det(A)\det(B)$. Since $X = PYP^{-1}$ we have $\det(X) = \det(PYP^{-1}) = \det(P) \det(Y) \det(P^{-1}) = \frac{\det(P)}{\det(P)} \det(Y) = \det(Y)$. Since $\det(P^{-1}) = \frac{1}{\det(P)}$

OCH
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