How would I go about showing that the inverse of a self-adjoint linear transformation on a finite dimensional real vector space is also self-adjoint?
By definition, $\displaystyle A^*:H\to H$ is the adjoint of $\displaystyle A:H\to H$ if $\displaystyle \langle Ax , y \rangle = \langle x , A^* y \rangle \forall x,y\in H$.
Now, $\displaystyle (A^{-1})^*:H\to H$ is the transformation for which $\displaystyle \langle A^{-1}x , y \rangle = \langle x , (A^{-1})^* y \rangle \forall x,y\in H$, which means
$\displaystyle \langle AA^{-1}x , y \rangle = \langle x , A^*(A^{-1})^* y \rangle \forall x,y\in H$, and therefore
$\displaystyle \langle x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H$.
The trick is to use this result to prove that $\displaystyle A(A^{-1})^*=I_{|H|}$. It follows that $\displaystyle A^{-1}=(A^{-1})^*$, which is to say that the inverse of $\displaystyle A$ is self-adjoint.
Possible alternative:
If M is a hermitian matrix of A, then
$\displaystyle B = {B^t} \Rightarrow {B^t}{B^{ - 1}} = I \Rightarrow {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}} \Rightarrow {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}$, and so the inverse of B is hermitian, and so the inverse of the linear transformation is self-adjoint.
Is that correct?
I don't see how $\displaystyle {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}}$ implies $\displaystyle {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}$, but then again I'm very rusty with this sort of thing.
The missing step from my proof is as follows:
$\displaystyle \langle x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H$
$\displaystyle \implies$ $\displaystyle \langle I_{|H|}x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H$
$\displaystyle \implies$ $\displaystyle A(A^{-1})^*=I_{|H|}^*=I_{|H|}$.