# Inverse of a self adjoint linear transformation

• Jan 8th 2010, 12:48 PM
laura9010
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?
• Jan 8th 2010, 01:33 PM
hatsoff
Quote:

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, $A^*:H\to H$ is the adjoint of $A:H\to H$ if $\langle Ax , y \rangle = \langle x , A^* y \rangle \forall x,y\in H$.

Now, $(A^{-1})^*:H\to H$ is the transformation for which $\langle A^{-1}x , y \rangle = \langle x , (A^{-1})^* y \rangle \forall x,y\in H$, which means

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

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

The trick is to use this result to prove that $A(A^{-1})^*=I_{|H|}$. It follows that $A^{-1}=(A^{-1})^*$, which is to say that the inverse of $A$ is self-adjoint.
• Jan 8th 2010, 01:39 PM
laura9010
Quote:

Originally Posted by hatsoff
By definition, $A^*:H\to H$ is the adjoint of $A:H\to H$ if $\langle Ax , y \rangle = \langle x , A^* y \rangle \forall x,y\in H$.

Now, $(A^{-1})^*:H\to H$ is the transformation for which $\langle A^{-1}x , y \rangle = \langle x , (A^{-1})^* y \rangle \forall x,y\in H$, which means

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

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

The trick is to use this result to prove that $A(A^{-1})^*=I_{|H|}$. It follows that $A^{-1}=(A^{-1})^*$, which is to say that the inverse of $A$ is self-adjoint.

Possible alternative:

If M is a hermitian matrix of A, then

$B = {B^t} \Rightarrow {B^t}{B^{ - 1}} = I \Rightarrow {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}} \Rightarrow {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}$, and so the inverse of B is hermitian, and so the inverse of the linear transformation is self-adjoint.

Is that correct?
• Jan 8th 2010, 02:57 PM
hatsoff
Quote:

Originally Posted by laura9010
Possible alternative:

If M is a hermitian matrix of A, then

$B = {B^t} \Rightarrow {B^t}{B^{ - 1}} = I \Rightarrow {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}} \Rightarrow {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}$, 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 ${B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}}$ implies ${B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}$, but then again I'm very rusty with this sort of thing.

The missing step from my proof is as follows:

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

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

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