#math/linear-algebra Suppose $T\in \mathcal{L}(V,W)$ and $u_{1}, \ldots{}, u_{n}$ and $v_{1}, \ldots{}, v_{n}$ are bases of $V$. Let $A=\mathcal{M}(T,(u_{1}, \ldots{}, u_{n}))$. Let $B=\mathcal{M}(T,(v_{1}, \ldots{}, v_{n}))$. Let $C=\mathcal{M}(I,(u_{1}, \ldots{}, u_{n}),(v_{1}, \ldots{}, v_{n}))$. Then $ A=C^{-1}BC $ $C$ is taking a vector in terms of $u_{1}, \ldots{}, u_{n}$ and spitting out an $n\text{-by-}1$ matrix of scalars with respect to $v_{1}, \ldots{}, v_{n}$. $B$ is performing the function $T$ with respect to basis $v_{1}, \ldots{}, v_{n}$ for input and for output. Then $C^{-1}$ takes a input with respect to $v_{1}, \ldots{}, v_{n}$ and spits out an $n\text{-by-}1$ matrix with respect to $u_{1}, \ldots{}, u_{n}$. $ \underbrace{ (\mathcal{U})\text{-coordinates} \longrightarrow (\mathcal{V})\text{-coordinates} }_{C} \longrightarrow \underbrace{\text{apply } T_{v}}_{B} \longrightarrow \underbrace{(\mathcal{V})\text{-coordinates} \longrightarrow (\mathcal{U})\text{-coordinates}.}_{C^{-1}} $ --- Suppose that $v_{1}, \ldots{}, v_{n}$ is a basis of $V$ and $T\in \mathcal{L}(V)$ is invertible. Then $\mathcal{M}(T^{-1})=(\mathcal{M}(T))^{-1}$ where both matrices are with respect to the basis $v_{1}, \ldots{}, v_{n}$. By Matrix product, we know We know... $(\mathcal{M}T){M}(T^{-1})=\mathcal{M}(I)$. And we know that $\mathcal{M}(I)=\mathcal{M}(T)\times \mathcal{M}(T)^{-1}$. Thus $\mathcal{M(T^{-1})}=\mathcal{M}(T)^{-1}$ by uniqueness of the matrix inverse.