I don't understand this $\left(\left[T\right]^B_C\right)^{-1}=\left[T^{-1}\right]^C_B$

Question

Let’s say we have $T:V \rightarrow V$ and $B$ and $C$ are bases of V

and $T$ has $T^{-1}$ so that $T^{-1}\left(T(\vec{v})\right)=\vec{v}$

i understand that $\left[T\right]_{C}^{B}=\left[\begin{array}{ccccc} | & | & | & | & |\\ \left[T(b_{1})\right]_{C} & \left[T(b_{2})\right]_{C} & \left[T(b_{3})\right]_{C} & \cdots & \left[T(b_{n})\right]_{C}\\ | & | & | & | & | \end{array}\right]$ where $B= \{ b_1,b_2,...,b_n\}$

and i get that $\left(\left[T^{-1}\right]_{B}^{C}\right)=\left[\begin{array}{ccccc} | & | & | & | & |\\ \left[\left(T^{-1}(c_{1})\right)\right]_{B} & \left[\left(T^{-1}(c_{2})\right)\right]_{B} & \left[\left(T^{-1}(c_{3})\right)\right]_{B} & \cdots & \left[\left(T^{-1}(c_{n})\right)\right]_{B}\\ | & | & | & | & | \end{array}\right]$

but i don't see the connection to $\left(\left[T\right]^B_C\right)^{-1}=\left[T^{-1}\right]^C_B$

I am especially confused by $C$ and $B$ flipping places

Answer 1

That's because\begin{align}[T]_B^C.[T^{-1}]_C^B&=[T.T^{-1}]_C^C\\&=[\operatorname{Id}]_C^C\\&=\operatorname{Id}.\end{align}Therefore,$$\left([T]_B^C\right)^{-1}=[T^{-1}]_C^B.$$