• Feb 4th 2010, 12:40 AM
typhoonss821
I recently teach myself linear algebra with Friedberg's textbook.
And I have a question about adjoint operator, which is on p.367.

Definition Let T : V → W be a linear transformation where V and W are finite-dimensional inner product spaces with inner products <‧,‧> and <‧,‧>' respectively. A funtion T* : W → V is called an adjoint of T if <T(x),y>' = <x,T*(x)> for all x in V and y in W.

Then ,my question is how to prove that there is a unique adjoint T* of T ?

Can anyone give me some tips ? thanks^^
• Feb 4th 2010, 02:39 AM
tonio
Quote:

Originally Posted by typhoonss821
I recently teach myself linear algebra with Friedberg's textbook.
And I have a question about adjoint operator, which is on p.367.

Definition Let T : V → W be a linear transformation where V and W are finite-dimensional inner product spaces with inner products <‧,‧> and <‧,‧>' respectively. A funtion T* : W → V is called an adjoint of T if <T(x),y>' = <x,T*(x)> for all x in V and y in W.

Then ,my question is how to prove that there is a unique adjoint T* of T ?

Can anyone give me some tips ? thanks^^

Suppose there are two adjoints, $\displaystyle T^*\,,\,\,T^{'}$ to a transf., then for all vectors we get : $\displaystyle <Tx,y>=<x,T^*y>=<x,T^{'}y>\Longrightarrow <x,(T^*-T^{'})(y)>= 0$ $\displaystyle \forall y\,,\,\,(T^*-T^{'})(y)=0\Longrightarrow T^*=T^{'}$

Tonio
• Feb 4th 2010, 08:01 PM
Roam
That was much simpler than what I was thinking: You know there are 4 main properties of the adjoint; Suppose S, T are linear operators on V and let $\displaystyle k \in K$ then

1. $\displaystyle (S+T)^*=S^*+T^*$
2. $\displaystyle (kT)^* = \bar{k}T^*$
3. $\displaystyle (ST)^*=T^*S^*$
4. $\displaystyle (T*)^*=T$

If you examine all of these properties, you will notice that the uniqueness of the adjoint implies that particiular property. For example $\displaystyle \forall u,v \in V$ , $\displaystyle < (S+T)(u), v > = <S(u) +T(u), v >= <S(u), v > + < T(u),v >$ $\displaystyle = < u, S^*(v) > + < u, T^*(v) > = < u, S^*(v) + T^*(v) > = <u, (S^*+T^*)(v) >$. You see the uniqueness of the adjoint implies (S+T)*=S*+T*.

Also, $\displaystyle <(kT)(u),v > = < kT(u), v > = k <u, T^*(v) >$ $\displaystyle = < u,\bar{k}T^*(v) > = <u, (\bar{k}T^*)(v)>$, again the uniqueness of the adjoint implies $\displaystyle (kT)^* = \bar{k}T^*$. The same with other two properties, they all show the uniqueness of the adjoint.