Does $P^{-1} A P = P P^{-1} A$?

Question

I hung up on the following step in a derivation I'm following and could use some guidance as to why it's true. I'm trying to show that the trace of a matrix is invariant under any similarity transformation, and I'm not sure why this step is legal.

\begin{align} \text{Tr} (\mathbf{B}) = \text{Tr} (\mathbf{P}^{-1}\mathbf{AP}) =\text{Tr} (\mathbf{PP}^{-1}\mathbf{A}) \end{align}

From there, it's simply the identity matrix and therefore the matrix $\mathbf A$ is invariant under similarity transformations. But, matricies are not commutative. So, I'm not sure the right hand side is true.

Answer 1

With Einstein's notation, we have that $$\text{Tr}(AB)=A_{ij}B_{ji}=B_{ji}A_{ij}=\text{Tr}(BA)$$ And for 3 matrices: $$\text{Tr}(ABC)=A_{ij}B_{jk}C_{ki}=C_{ki}A_{ij}B_{jk}=\text{Tr}(CAB)$$

Answer 2

It is true because, for any two $n\times n$ matrices $A$ and $B$, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$.

Answer 3

Let $U=(U_{ij})_{n\times n}$ and $V=(V_{ij})_{n\times n}$ such that $V=P^{-1}$ and $U=P$. We have \begin{align} UAV =& (U_{ij})_{n\times n} \cdot (A_{ij})_{n\times n} \cdot (V_{ij})_{n\times n} \\ =& \left(\sum_{k=1}U_{ik}\cdot A_{kj}\right)_{n\times n} \cdot (V_{ij})_{n\times n} \\ =& \left(\sum_{\ell=1}^{n}\big(\sum_{k=1}^nU_{ik}\cdot A_{k\ell}\big)\cdot V_{\ell j}\right)_{n\times n} \\ =& \left(\sum_{\ell=1}^{n}\sum_{k=1}^nU_{ik}\cdot A_{k\ell}\cdot V_{\ell j}\right)_{n\times n} \end{align} Then $U=P$ and $V=P^{-1}$ implies $UV=(U_{i\ell})_{n\times n}\cdot(V_{i\ell})_{n\times n}=(\sum_{k=1}^{n}U_{ik}V_{k\ell})_{n\times n}=I$. More explicitly, $$ \sum_{k=1}^{n}U_{ik}V_{k\ell}=\begin{cases} 1 & \mbox{if } i=\ell\\ 0 & \mbox{if } i\neq \ell\end{cases} $$ With this in mind we obtain the identitie below \begin{align} \mbox{trace}(UAV) =& \sum_{i=1}^{n} \sum_{\ell=1}^{n}\sum_{k=1}^n U_{ik}\cdot A_{k\ell}\cdot V_{\ell i} \\ =& \sum_{k=1}^n \sum_{\ell=1}^{n} \Big(\sum_{i=1}^{n} U_{ik}\cdot V_{\ell i}\Big)\cdot A_{k\ell} \\ =& \sum_{k=1}^n A_{k k} \end{align} In an entirely analogous way \begin{align} UVA =& (U_{ij})_{n\times n} \cdot (V_{ij})_{n\times n} \cdot (A_{ij})_{n\times n} \\ =& \left( \sum_{r=1}U_{ir}\cdot V_{rj}\right)_{n\times n} \cdot (A_{ij} )_{n\times n} \\ =& \left( \sum_{s=1}^{n}\big(\sum_{r=1}^n U_{ir}\cdot V_{rs}\big)\cdot A_{s j} \right)_{n\times n} \\ =& \left( \sum_{s=1}^{n}\sum_{r=1}^n U_{ir}\cdot V_{rs}\cdot A_{s j} \right)_{n\times n} \end{align} implies $$ \mbox{trace}(UVA)= \sum_{t=1}^{n}\sum_{s=1}^{n}\sum_{r=1}^n U_{tr}\cdot V_{rs}\cdot A_{st} =\sum_{t=1}^{n} A_{tt} $$