1. ## Inverse of a self adjoint linear transformation

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?

2. Originally Posted by laura9010
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.

3. Originally Posted by hatsoff
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?

4. Originally Posted by laura9010
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:

$\displaystyle \langle x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H$

$\displaystyle \implies$ $\displaystyle \langle I_{|H|}x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H$

$\displaystyle \implies$ $\displaystyle A(A^{-1})^*=I_{|H|}^*=I_{|H|}$.
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