Is the Adjoint of an Invertible Operator Also Invertible?

[redyelloworange]

Homework Statement

Let V be a finite-dimensional inner product space, and let T be a linear operator on V. Prove that if T is invertible, then T* is invertible and (T*)^-1 = (T^-1)*

Homework Equations

As shown above.
<T(x),y> = <x,T*(y)>

The Attempt at a Solution

Well, I figure you only need to show that the equation holds, that shows that T* is invertible, since its inverse exists.
Now, I try to do something with the inner products:

<(T^-1)(x),y> = <x,(T^-1)*(y)>

I’m not sure how to “flip” inverse and the star.

Thanks for your help! =)

 

[AKG]

Start with <T*(T^-1)*x,y> and see what you end up with.

 

[redyelloworange]

<T*(T^-1)*x,y> = <T*(x), T^-1(y)> = <(T^-1)*x,T(y)> = <x, (T^-1)(T)y> = <x,y>

so T*(T^-1)* = I, so (T^-1)* is the inverse of T*, hence (T^-1)* = (T*)^-1)

thanks! =)

 

[AKG]

Step 1 is to deduce that <T*(T^-1)*x,y> = <x,y>. Once you get that, step 2 is exactly what you did:

T*(T^-1)* = I, so (T^-1)* is the inverse of T*, hence (T^-1)* = (T*)^-1

But you appear to have done step 1 wrong. You made the right conclusion, but all your steps look invalid. For example, you first line is of the form <A*B*x,y> = <A*x,By>. Why is this wrong in general? Well if it were always true, we'd get:

<A*B*x,y> = <A*x,By>
<A*B*x,y> = <B*A*x,y>
<(A*B* - B*A*)x,y> = 0 (for all y)
A*B* - B*A* = 0
AB = BA (in general, i.e. for all A and B)

But matrix multiplication is not commutative in general, so this is wrong. Try again.

 

[juaninf]

then how is?

 

[HallsofIvy]

Then do it being very careful about order.

<A*B*x,y> = <B*x,Ay>= <x, BAy>.

<B*A*x,y>= <A*x, By>= <x, ABy>.

 

[juaninf]

Understand but, In first prove (T* is invertivel) but how sure that <T*(T-1)*x,y> = <x,y> , if i don't know that (TT-1) is I, because is just that want will prove