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Let $V$ be a finite dimensional vector space over an arbitrary field $F$ and let $S$ be an ordered basis of $V$. Let $A$ be an operator from $V$ to $V$. How can we prove that the matrix of $A$ with respect to $S$ is invertible, ie how to show that $M_{A^{-1}} = {M_A}^{-1}$?

user64799
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1 Answers1

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By definition of invertibility, there is an operator $A^{-1}$ such that $A\circ A^{-1}=A^{-1}\circ A=Id$, the identity transformation.

Both of them have matrix representations $M_A$ and $M_{A^{-1}}$, as well as $M_{Id}=I_n$. Since matrix multiplication by these matrices matches the action of the transformations, the equation $A\circ A^{-1}=A^{-1}\circ A=Id$ translates into $M_A M_{A^{-1}}=M_{A^{-1}}M_A=I_n$.

rschwieb
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  • The final step seems a little non rigorous to me; is there any way you could prove $M_{A^{-1}} = {M_A}^{-1}$ instead? – user64799 Apr 19 '13 at 19:28
  • The last equation says exactly $M_A M_{A^{-1}}=M_{A^{-1}}M_A=I_n $. There's no rigor missing: remember that $B(x)=M_B\cdot x^T$ for all $x$, so there's no doubt the equations hold. – rschwieb Apr 19 '13 at 19:44
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    @user64799 This is as rigorous as can be, +1. You should check more generally for yourself that when a basis is fixed, the mapping $A\mapsto M_A$, which takes an operator $A\in L(V)$ to its matrix $M_A\in M_n(F)$ with respect to that basis, is an isomorphism of $F$-algebras. A key step is to verify that $M_{A\circ B}=M_AM_B$. Then the inverse property you want follows like rschwieb said. – Julien Apr 19 '13 at 20:46