if $T$ is isomorphism, how can I prove that $[T^{-1}]_B=[T]_B^{-1}$ for any base $B$ of $V$?

Question

given vector space $V$ so that $\dim(V)$ is finite , and linear-transformation $T:V \to V$.

if $T$ is isomorphism, how can I prove that $[T^{-1}]_B=[T]_B^{-1}$ for any base $B$ of $V$?

Answer 1

Using thae fact that$$[T]_B.[T^{-1}]_B=[T.T^{-1}]_B=\operatorname{Id}.$$

Answer 2

Use the definition of $[T]_B$. Remember that if $[v]_B$ is the (column) vector whose entries are the coordinates of $v$ in the base $B$, then $[T]_B$ is the $n\times n$ matrix such that $$[T]_B\cdot [v]_B=[T(v)]_B, \quad \forall v\in V,$$ that is the (only) matrix such that when it premultiplies $[v]_B$ gives a (column) vector whose entries are the coordinates of $T(v) \in V$ in the base $B$.

Now, what would the matrix $[T^{-1}]_B$ be?

Answer 3

$T\circ T^{-1}=I_V$, so $[T]_B [ T^{-1}]_B=I$·