Proof of Subspace of Self-Adjoint Operators in L(V)

[evilpostingmong]

Homework Statement

Show that if V is a real inner-product space, then the set
of self-adjoint operators on V is a subspace of L(V).

Homework Equations

The Attempt at a Solution

Let M be the matrix representing T. Since we are dealing with real numbers, and T is self-adjoint, T=T* so M=M^T.
Let u and w be vectors. Now <Mu, w>=<u, M^Tw>. Since M=M^T, <u, M^Tw>=<u, Mw>.
If M=M^T, M is an nxn matrix. And since M and M^T are nxn, M and M^T map a vector from an n-dimensional space V back to an n-dimensional space V. Since nxn matrices can only multiply into n-tall vectors, u and w must be n-tall vectors. Therefore u and w must be in V. Since M and M^T are nxn matrices and can map an n-tall vector from V to an n-dimensional space (V), the set of self adjoint operators (which M and M^T are a part of)are within the set of L(V).

 

[Office_Shredder]

The set of self adjoint operators is obviously contained within L(V) by definition of self adjoint... you don't need to do that matrix mumbo jumbo to prove it. What the question is asking is to show it's a subspace which you haven't done

 

[Office_Shredder]

Look back at your post. Did you prove:
1) 0 is self adjoint
2) If T and S are self adjoint, T+S is self adjoint
3) If T is self adjoint, a real, then aT is self adjoint.

It looks like you tried to prove T* is linear or something. That's not what you need to do. L(V) is a vector space of linear functions from V to V. Show the set of self adjoint linear functions is a vector subspace

 

[evilpostingmong]

Look back at your post. Did you prove:
1) 0 is self adjoint
2) If T and S are self adjoint, T+S is self adjoint
3) If T is self adjoint, a real, then aT is self adjoint.

It looks like you tried to prove T* is linear or something. That's not what you need to do. L(V) is a vector space of linear functions from V to V. Show the set of self adjoint linear functions is a vector subspace

Ok Let w and k be vectors in V. To show 0 is a self adjoint transformation, <0w, k>=<w, 0k>=<0,k>=<w, 0>=0. Since <w, 0k>=<0w, k>, 0 is self adjoint.

Now let T and S be self adjoint so that <w, Tk>=<Tw, k> and <w, Sk>=<Sw, k>.
Now <(S+T)w, k>=<Sw+Tw, k>. If S*=S and T*=T, S*+T*=S+T therefore
S+T=(S+T)*. So S+T is self adjoint

I'll do the third, just want to know if I have it here.
It kinda gets me how you can span make a span out of operators. I mean unless you
represent operators as matrices and put those in a span.

 

[Office_Shredder]

Have you proven that S*+T*=(S+T)*? Likely what's expected is something more like

<(S+T)w,k> = <Sk+Tk,w> = <Sk,w> + <Tk,w> = <k,Sw> + <k,Tw> = <k,Sw+Tw> = <k,(S+T)w>

but assuming you have S*+T* = (S+T)* then you're correct

 

[evilpostingmong]

Have you proven that S*+T*=(S+T)*? Likely what's expected is something more like

<(S+T)w,k> = <Sk+Tk,w> = <Sk,w> + <Tk,w> = <k,Sw> + <k,Tw> = <k,Sw+Tw> = <k,(S+T)w>

but assuming you have S*+T* = (S+T)* then you're correct

Well yeah I figured that since S is self adjoint and T is self adjoint, and were not
dealing with the complex space, S*=S and T*=T. So S*+T*=S+T because
S=S* and T=T*. And (S+T)* =S*+T* is an identity, I apologize, as I didn't know I
actually had to prove (S+T)*=S*+T* since I didn't think that it the main aim of the proof.
I'm doing the third one.

 

[Office_Shredder]

Even thought S*+T* = S+T, without knowing that S*+T*=(S+T)* you can't conclude (S+T)*=S+T

 

[evilpostingmong]

3) If T is self adjoint, a real, then aT is self adjoint.
Now using w and k in V, we have <Tw, k>=<w, Tk>=<T*w, k>=<w, T*k>.
Now <aw, k>=<w, ak>=a<w,k>. If <Tw, k>=<w,Tk> then a<Tw, k>=a<w, Tk>.
Or <aTw,k>=<w,aTk>.

 

[evilpostingmong]

am i right?

 

[Office_Shredder]

Looks good to me

 

[evilpostingmong]

Thanks! But here's part two. This one is similar to the first.

Show that if V is a complex inner-product space, then the
set of self-adjoint operators on V is not a subspace of
L(V).

I think that this has to do with the fact that say A is a complex matrix rep. T
and A* is a complex matrix rep. T*. A* is a complex conjugate=/=A.
But the problem is that the set needs to contain operators that are self-adjoint.
So I guess that A is a real matrix. But since the set of real adjoint operators
cannot be within L(V), since V is a complex space, the set cannot be a
subspace of L(V).

After I get this out of the way, I'll prove this statement.

 

[Office_Shredder]

Don't fall back on matrix representation arguments. Try to complete the same steps as you did for the real case. One of them won't work, and it will be because you're in a complex space instead of a real space.

 

[evilpostingmong]

Don't fall back on matrix representation arguments. Try to complete the same steps as you did for the real case. One of them won't work, and it will be because you're in a complex space instead of a real space.

lol I need to go to matriholics anonymous. alright that was the worst joke ever posted
in the history of Physics Forums

Ok
prove 0 is self adjoint

let k+wi and j+hi be vectors in V. Now <0(k+wi), j+hi>=<k+wi, 0(k+wi)>=<0,j+hi>
=<k+wi, 0>=0.
works

prove T+S is self adjoint where T is a real transformation and S is a complex transformation.

Using the same vectors k+wi and j+hi we have <T(k+wi), j+hi>=<k+wi, T*(j+hi)>
and <S(k+wi), j+hi>=<k+wi, S*(j+hi)>. Now since T is a real
transformation, it is self adjoint so it is within the set of self adjoint transformations
in L(V). But since L(V) is complex, S is also in L(V). And S*(k+wi) is the complex conjugate
of S(k+wi). Therefore S=/=S*. Since S=/=S*, T+S=T*+S but T+S=/=(T+S)*
To show this consider <(T(k+wi), j+hi>+<S(k+wi), j+hi>=<k+wi, T*(j+hi)>+<k+wi, S*(j+hi)>. Since T*=T, but S*=/=S, we have <(T(k+wi), j+hi>+<S(k+wi), j+hi>=<k+wi, T(j+hi)>+<k+wi, S*(j+hi)> or <(T+S)(k+wi), j+hi>=<k+wi, (T+S*)(j+hi)>. This shows that (T+S)*=T*+S*=T+S* but (T+S)*=/=T+S.

 

[Office_Shredder]

Ummm... close. You almost hit upon the right point, but your proof if far more complicated than it needs to be. If you have (T+S)*=T*+S*, and you assumed T and S are self adjoint, then (T+S)*=T+S./ You got yourself confused trying to introduce complex numbers, but those are dealt with in the proof for closure under scalar multiplication. Start there

 

[evilpostingmong]

Ummm... close. You almost hit upon the right point, but your proof if far more complicated than it needs to be. If you have (T+S)*=T*+S*, and you assumed T and S are self adjoint, then (T+S)*=T+S./ You got yourself confused trying to introduce complex numbers, but those are dealt with in the proof for closure under scalar multiplication. Start there

Ok I'm pretty sure the closure under scalar multiplication would
be far simpler.

Prove aT is self adjoint.

Let k+wi and j+hi be vectors in V. Let T be a self adjoint operator. Now <T(k+wi), j+hi>=<k+wi, T(j+hi)>. Let a=x+yi a scalar. Now a<T(k+wi), j+hi>=<aT(k+wi), j+hi>. Now we have <(aT)(k+wi), j+hi>. Since aT is complex, a*T* is the complex conjugate of aT, so aT=/=a*T*. Thus <(aT)(k+wi), j+hi>=<k+wi, (a*T*)(j+hi)>=/=<k+wi, (aT)(j+hi)> since
a=x+yi and a*=x-yi. Thus a<T(k+wi), j+hi> is not self adjoint which disproves closure.

 

[Office_Shredder]

What is this k+wi stuff? Why can't you just pick a vector?

If T is self adjoint then

<aTu,w> = a<Tu,w> = <u,Tw> = <u,a*Tw> where a* is the conjugate of a. See where the problem is?

 

[evilpostingmong]

What is this k+wi stuff? Why can't you just pick a vector?

If T is self adjoint thend

<aTu,w> = a<Tu,w> = <u,Tw> = <u,a*Tw> where a* is the conjugate of a. See where the problem is?

k+wi is a complex vector. That proof you have is pretty much the same as mine though I
admit I've done some unnecessary things. I determined that aT=/=(aT)* since a*=/=a
so a*T*=/=aT There is a complex scalar a that when multiplied by T produces a complex
transformation (aT). So the fact that a exists prevents the set of self adjoints to be closed. But again,
this isn't the most elegant way to do it.