• fluidistic

Gold Member

## Homework Statement

I must show several properties about linear operators using the definition of the adjoint operator.
A and B are linear operator and ##\alpha## is a complex number.
The first relation I must show is ##(\alpha A + B)^*=\overline \alpha A^*+B^*##.

## Homework Equations

The definition I have an an adjoint is: ##A^*## is the adjoint of ##A## if ##\langle g,Af \rangle = \langle A^* g ,f \rangle## where f and g are any vectors in a Hilbert space.

## The Attempt at a Solution

Let ##C^*=(\alpha A+B)^*##. Using the definition of adjoint I get: ##\langle C^*g,f \rangle=\langle g, Cf \rangle \Rightarrow \langle (\alpha A+B)^*g ,f \rangle =#### \langle g, (\alpha A+B)f \rangle =\langle g, \alpha A \cdot f \rangle + \langle g, Bf \rangle = \alpha \langle g, Af \rangle + \langle g ,Bf \rangle = \alpha \langle A^*g, f \rangle + \langle B^*g ,f \rangle##.
But I'm getting lost. I've no idea how I can obtain A, B, A^* and B^* using the definition of the adjoint.

Oh wait, on my draft I think I have finished the "proof". The last expression is worth ##\langle \overline \alpha A^*g ,f \rangle + \langle B^* g, f \rangle = \langle (\overline \alpha A^* + B^*)g ,f \rangle##. Then by associativity ##(\alpha A+B)^*=\overline \alpha A^* + B^*##.
Does this look right?

OK, so you proved

$$<(\alpha A + B)^* g,f> = <(\overline{\alpha}A^* + B^*)g,f>$$

But how do you obtain from there that ##(\alpha A + B)^* = \overline{\alpha}A^* + B^*##?

OK, so you proved

$$<(\alpha A + B)^* g,f> = <(\overline{\alpha}A^* + B^*)g,f>$$

But how do you obtain from there that ##(\alpha A + B)^* = \overline{\alpha}A^* + B^*##?
Good question, I don't know. To simplify the notation I must show that if ##\langle Dg, f \rangle = \langle Eg,f \rangle## then D=E, where D and E are linear operators; for any inner product in a Hilbert space.
When I look at the definition of the inner product, I only see the positive definiteness, linearity and conjugate symmetry. I don't see how that would help me.

Something that I think would help is the axiom ##<x,x>=0## then ##x=0##. Because this is the only axiom that allows you to begin with the inner product and end with a statement about a vector.

Let's prove a more general thing. Let x,y be vectors such that ##<x,z> = <y,z>## holds for all z, then x=y.

To prove this, note that the above is equivalent to saying ##<x-y,z>=0## for all z. Does this give you a hint?

I'm not 100% sure.
Here is my attempt: ##\langle x ,z \rangle =\langle y, z \rangle## holds for all z.
I add "-y" in the first argument of each side (can I really do that without demonstrating that the equality still hold?) to get ##\langle x-y,z \rangle =\langle y-y ,z \rangle = \langle 0 ,z \rangle =0##
So that ##\langle x-y, z \rangle =0## must hold for all z. This implies that ##x-y=0##. End of proof.

P.S.:Yes I can add -y in the first argument of the inner product. To prove it is simple, using the linearity of the inner product.

So that ##\langle x-y, z \rangle =0## must hold for all z. This implies that ##x-y=0##.

This implication is not very clear. why does ##<x-y,z>=0## for all z imply that ##x-y=0##?

It is of utmost importance to mention that the operators are bounded and defined everywhere on the Hilbert space, else one needs a new reasoning.

This implication is not very clear. why does ##<x-y,z>=0## for all z imply that ##x-y=0##?

Because in particular ##z=x-y##. So we have ##\langle x-y , x-y \rangle =0##, which is possible if and only if ##x-y=0##. Thus ##x=y##. End of proof?
dextercioby said:
It is of utmost importance to mention that the operators are bounded and defined everywhere on the Hilbert space, else one needs a new reasoning.
Is is the same as showing that the operators are continuous over the whole Hilbert space?

Because in particular ##z=x-y##. So we have ##\langle x-y , x-y \rangle =0##, which is possible if and only if ##x-y=0##. Thus ##x=y##. End of proof?[...]

Yes, good point.

[...]Is is the same as showing that the operators are continuous over the whole Hilbert space?

No, continuity is an alternative assumption to boundedness, because for ∞-dim separable Hilbert spaces the 2 notions are equivalent.

There is a connection between continuity (and equivalent: boundedness) and being everywhere defined.

A linear operator ##T:D\rightarrow H## is usually defined on a dense subset ##D## of the hilbert space ##H##. Now, it turns out that if ##T## is continuous on ##D##, then there exists a unique extension of ##T## on the entire Hilbert space H. This is called the BLT theorem (BLT = bounded linear transformation). So if ##T## is continuous on a dense subset, then we can make it everywhere defined.

If ##D## is not dense, then we can only extend ##T## on ##\overline{D}##.