Proof of equal characteristic polynomials for congruent matrices?

Question

It is probably a very elementary question but I cannot figure it out: my book gives a proof for the theorem that congruent matrices have the same determinant, characteristic polynomial ($\phi(X)$) and the same trace. However, I do not understand the second and third operation they conduct in the proof. How do they go from $(X \mathbb{I_n} - P^{-1}\cdot A \cdot P)$ to $(P^{-1}\cdot (X \mathbb{I_n} - A) \cdot P)$ and why is $det(P^{-1})\cdot\det(X \mathbb{I_n} - A) \cdot det(P)$ equal to $\phi_A(X)$?

Many thanks for your help in advance!

Consider two matrices $A$ and $B$ ($\in \mathbb{R^{m \times n}}$) to be congruent. If there is an invertible matrix $P$, such that $B = P^{-1} \cdot A \cdot P$, then: \begin{equation} \begin{aligned} \phi_B(X) &= det(X \mathbb{I_n} - P^{-1}\cdot A \cdot P) \\ &= det(P^{-1}\cdot (X \mathbb{I_n} - A) \cdot P) \\ &= det(P^{-1})\cdot\det(X \mathbb{I_n} - A) \cdot det(P) \\ &= \phi_A(X) \end{aligned} \end{equation}

Answer 1

For the second step: use that $\det(XY)=\det(X)\det(Y)$ for all $X,Y$, so that we have $\det(P^{-1})\det(P)=det(P^{-1}P)=\det(I_n)=1$. Since $K$ is commutative, and $\det(X)\in K$, you can interchange the determinants, i.e., $$ det(P^{-1})\cdot\det(X \mathbb{I_n} - A) \cdot det(P)=\det(X \mathbb{I_n} - A) \cdot det(P^{-1})\det(P)=\det(X \mathbb{I_n} - A)=\phi_A(X). $$ The first step follows similarly.