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^^

2. 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, $T^*\,,\,\,T^{'}$ to a transf., then for all vectors we get : $==\Longrightarrow = 0$ $\forall y\,,\,\,(T^*-T^{'})(y)=0\Longrightarrow T^*=T^{'}$

Tonio

3. 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 $k \in K$ then

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

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

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