Definition from my textbook: For each linear operator T on a inner product space V, the adjoint of T is the mapping T* of V into V that is defined by the equation <T*(v),w> = <v,T(w)> for all v, w E V.

My instructor defined it by <T(v),w> = <v,T*(w)> and he said that these 2 definitions are equivalent.

Now, can someone please explain WHY they are equaivalent?

Can both definitions be used at the SAME time, or do I have to choose 1 of the 2 definitions and use this chosen definition consistently everywhere?


Thanks for explaining!

Yes,, they are exactly equivalent. To see this, merely note that (T*)* = T. Can you see why? Have you done representations of linear operators yet? When you do, which should be shortly, it becomes obvious. Your instructor's form is the more usual, but they are exactly the same!

Definition: A linear operator is self-adjoint if and only if T*=T

But not all adjoint operators are self-adjoint, right?

[I've done represetntations of linear operators by matrices]

No, not all operators are self-adjoint, this is true. But if T is and operator, and T* is it's adjoint, then the adjoint of T* is (T*)* = T. Look at the matrix representations and you'll see why!

I see now what you were asking. (Tv,w) = (T*w) and (T*w,v) = (w,TV) are equivalent as definitions. They clearly give numerically different results, which my comment about "exactly the same" might have led you to think. Sorry if I confused you.

OK, maybe you'd prefer this:

By the inner product axioms, (v,w) = \overline{(w,v)} (complex conjugate)

(v,Tw) = \overline{(Tw,v)} = \overline{(w,T\dagger v)} = (T\dagger v,w)

Just change the roles of w and v then you would get the other definition.

Thanks!

So if I am solving a problem, can both definitions be used at the SAME time within the same problem, or do I have to choose 1 of the 2 definitions and use this chosen definition consistently everywhere?

Say, for example, if we consider C^2 with the standard inner product and T: C^2->C^2 is defined by T(x,y)=(x+(1-i)y, (1+i)x+2y), how exactly can I find T*(x,y)?

The definition of T* seems obscure to me...

Please double check my calculation: We have

<T(x,y),(z,t)> = <(x+(1-i)y, (1+i)x+2y),(z,t)> = xz*+(1-i)yz*+(1+i)xt*+2yt*
= x(z*+(1+i)t*) + y[(1-i)z*+2t*]=x(z+(1-i)t)* + y[(1+i)z+2t]*
= <(x,y),(z+(1-i)t,(1+i)z+2t)>.

This should be equal to <(x,y),T*(z,t)>, so

T*(x,y) = (x+(1-i)y,(1+i)x+2y),

which shows that T is Hermitian.

Thanks a lot!

But what is the relation between (z,t) and (x,y)? The question is looking for T*(x,y), but in your calculations, it seems that (z,t) is used instead of (x,y). I get a little confused here...

It is just I used (x,y) instead of (z,t) in the last line. It is like f(x)=x+3 and f(z)=z+3 are two ways to write the same function. If you want, you can write as T*(z,t) = (z+(1-i)t,(1+i)z+2t).

Thanks!

I have one more question about inner product.

Let u=(u1,u2), v=(v1,v2)
I denote (u1)* as conjugate of u1

How can I prove that <u,v>=2(u1)*v1 + (1-i)(u1)*v2 + (1+i)(u2)*v1 + 3(u2)*v2 defines an inner product on C2?

Now
[2 1-i
1+i 3]
is a Hermitian matrix, so if I can prove that <v,v> >0 and <v,v>=0 iff v=0, then I'm done, but how can I do so?

Write the expression for <v,v> explicitly, and then use the inequality 2ab<=a^2+b^2 to bound the expression from below.

<v,v>=2(v1)*v1 + (1-i)(v1)*v2 + (1+i)(v2)*v1 + 3(v2)*v2

(v1)*v2 and (v2)*v1 won't cancel, so how can the i's cancel?

They will not cancel, but you can expand the complex numbers into real and imaginary parts and then use the above inequality. They may be faster ways but I cannot think of one.