Let $f$ be a bijective linear transformation. Why does it hold that $A_{f^{-1}} = A^{-1}_f$, where $A_f$ denotes the transformation matrix corresponding to $f$? I can apply the theorem and understand what it says. What I'm looking for is a proof. So far, I understand the following:
- A function has an inverse iff it is bijective.
- Every linear transformation $f: V \rightarrow W, x \mapsto f(x)$ with arbitrary vector spaces $V$ and $W$ is defined by a transformation matrix $A$ such that $f: V \rightarrow W, x \mapsto Ax$.
I couldn't find a proof in my lecture notes and don't know where to start doing it myself so any help would be appreciated.
Best Answer
Fix a a basis $B$ of a finite-dimensional vector space $V$. For each linear map $f\colon V\longrightarrow V$, let $A_f$ be the matrix of $f$ with respect to the basis $B$. Then, if $g\colon V\longrightarrow V$ is another linear map, you have$$A_{g\circ f}=A_gA_f\quad\text{and}\quad A_{f\circ g}=A_fA_g.$$So, in particular,$$\operatorname{Id}=A_{f^{-1}\circ f}=A_{f^{-1}}A_f\quad\text{and}\quad \operatorname{Id}=A_{f\circ f^{-1}}=A_fA_{f^{-1}}.$$That is$$A_{f^{-1}}=A_f^{\,-1}.$$