Why is the characteristic polynomial of a linear map, $f$, independent of the basis w.r.t. which the matrix $A$ represents f?

Question

In the book that I'm reading it states that:

It is given that the linear map $f$ is such that $f:V\rightarrow V$ and that its characteristic polynomial is given by:

$$\chi_A(\lambda)=\det(A-\lambda I) $$

where $A$ is the matrix that represents $f$ w.r.t. some basis

$$\chi_{A'}=\chi_{PAP^{-1}}(\lambda)=\det[P(A-\lambda I)P^{-1}]=\det(A-\lambda I)= \chi_A(\lambda) \qquad [1]$$

where $A'$ is a matrix representing $f$ w.r.t. a different basis.

However, I don't understand the equation [1], before seeing it I was expecting an equation of the form

$$\chi_{A'}=\chi_{PAP^{-1}}(\lambda)=\det[PAP^{-1}-\lambda I] \qquad [2] $$

I'm not sure how to continue [2]; neither do I understand how any of the steps of equations [1] are related to one another. I'd appreciate it if you would help me out.

Answer 1

Note that $\lambda I$ is a matrix that commutes with every matrix. So $$P(A-\lambda I)P^{-1} = PAP^{-1} - P(\lambda I)P^{-1} = PAP^{-1} - (\lambda I)PP^{-1} = PAP^{-1}-\lambda I.$$

So from (2) you can get to (1). Once you get to $1$, you are using the fact that $$\det(XY) = \det(X)\det(Y) = \det(Y)\det(X) = \det(YX)$$ for any two matrices $X$ and $Y$. Thus, $$\begin{align*} \det(P(A-\lambda I)P^{-1}) &= \det(P)\det(A-\lambda I)\det(P^{-1}) \\ &= \det(P)\det(P^{-1})\det(A-\lambda I)\\ &= \det(A-\lambda I) \end{align*}$$ because $\det(P)\det(P^{-1}) = 1$.

Answer 2

In $[1]$ they simply use the multiplicativity of the determinant; $$\det(PAP^{-1}-\lambda I)=\det\bigl(P(A-\lambda I)P^{-1}\bigr)=\det P\,\det(A-\lambda I\,\det(P^{-1})=\det P\,\det(A-\lambda I)\,(\det P)^{-1}.$$