Normal operators and self adjointness

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evilpostingmong
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Homework Statement


Suppose V is a complex inner-product space and T ∈ L(V) is a
normal operator such that T9 = T8. Prove that T is self-adjoint
and T2 = T.

Homework Equations


The Attempt at a Solution


Consider T9=T8. Now "factor out" T7 on both sides to get T7T2 =TT7. Now we represent T as a matrix. Since T is normal, it is diagonizable. Therefore it is invertible. Now T2=TT7T-7 so T2=(T)(I)7=T.
 
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Dick said:
The zero matrix is diagonalizable. It's not invertible. And it's not the only such matrix.

oh right. Can't forget those.
Let T be a "reduced" matrix. Now let T be a zero transformation. It is clear that T2=T when T=0. Consider T=/=0 (still a "reduced" matrix). Consider T9=T8. Now "factor out" T7 on both sides to get T7T2 =TT7. Now we haveT7T2-TT7=0. Or
T7(T2-T)=0. Since we know that this equation holds true (equals 0), T2 must=T.
 
That's far from a valid proof, T^7 can be 0,even if it's not, that doesn't imply T^2-T=0.
 
I'm really stuck here. I mean, is it possible to prove T2=T
without proving T is self adjoint first? I kept coming up with ways to prove T2=T.
but they all failed because of the fact that Tn could be 0.
Also there are many cases where T is normal, and T9=T8 leads to
T2=T. I mean, T could be an identity, T could be a 0 map,
T could be an orthogonal projection, let me know if I missed something, lol.
Sorry but I can't think of a proof for this, if there is, a small hint could do the trick.
Oh and I COULD say take T-7 and multiply it by both sides. But
what if T7 is 0? So that's out of the question.
 
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The definition of "normal" is that TTT= TTT. I don't see where you have used that anywhere.

Also, are you allowed to assume that the vector space is finite dimensional? You seem to be doing that.
 
HallsofIvy said:
The definition of "normal" is that TTT= TTT. I don't see where you have used that anywhere.

Also, are you allowed to assume that the vector space is finite dimensional? You seem to be doing that.

I'm not sure whether or not I'm allowed to assume the space is finite dimensional.
 
evilpostingmong said:
T could be an identity, T could be a 0 map,
T could be an orthogonal projection, let me know if I missed something, lol.
(Note that the identity and the zero map are orthogonal projections) Do you think these are all of the cases? Can you prove it?


But what if T7 is 0?
Yes... what if...

(did you mean "singular" rather than "zero"?)
 
HallsofIvy said:
The definition of "normal" is that TTT= TTT. I don't see where you have used that anywhere.
I thought he invoked that to claim the operator was diagonalizable. (I confess I don't know the relevant theorem)
 
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Hurkyl said:
I thought he invoked that to claim the operator was diagonalizable. (I confess I don't know the irelevant theorem)

you're right on that.
 
Why don't you try putting T^9 and T^8 into diagonal form and see what it implies?
P.S.:Normal matrix is not just diagonalizable, also it can be diagonalized by unitary matrix.
 
kof9595995 said:
Why don't you try putting T^9 and T^8 into diagonal form and see what it implies?
P.S.:Normal matrix is not just diagonalizable, also it can be diagonalized by unitary matrix.
That is what I meant by reduced. Diagonal form (ie if singular, all 0's).
 
Hurkyl said:
That's not correct...

well ok but when I said reduced I meant diagonal form. Like when you reduce a singular matrix,
you may get all 0's on the matrix.And when you reduce a non-singular matrix, you get a matrix with all 1's across the diagonal
and 0's everywhere else.
 
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evilpostingmong said:
Like when you reduce a singular matrix,
you may get all 0's on the matrix.And when you reduce a non-singular matrix, you get a matrix with all 1's across the diagonal
and 0's everywhere else.
Where did you get this idea? It's not correct in most cases, but lucky you it's just the case in your problem.
 
All that you know so far is that the transformation has a basis in which it's diagonal, since it's normal. Now what does T^9=T^8 tell you about the diagonal elements?
 
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Dick said:
All is that you know so far is that the transformation has a basis in which it's diagonal, since it's normal. Now what does T^9=T^8 tell you about the diagonal elements?

That all of them are 0 along the diagonal, some but not all are 0 (rest are 1) along
the diagonal, or every element is 1 along the diagonal.
 
evilpostingmong said:
That all of them are 0 along the diagonal, some but not all are 0 (rest are 1) along
the diagonal, or every element is 1 along the diagonal.

In other words, every diagonal element is either 0 or 1. That's ok. But how do you know that? We are trying to do a proof here, right? Not an opinion poll.
 
Dick said:
In other words, some are 1 and some are 0. That's ok. But how do you know that? We are trying to do a proof here, right?
Roger. Now we know that T9=T8. Now is it okay
if I assume that "T" is reduced all the way to diagonal form? Because that
would leave us with the three types I mentioned earlier where T has all 0's, some
1's and some 0's, or all 1's along its diagonal, and 0's everywhere else (a characteristic
shared by all 3 cases). Then I can split the proof in two: one case where T is 0, the other where
T has at least one "1" on its diagonal. And life would be a lot easier.
 
evilpostingmong said:
Roger. Now we know that T9=T8. Now is it okay
if I assume that "T" is reduced all the way to diagonal form? Because that
would leave us with the three types I mentioned earlier where T has all 0's, some
1's and some 0's, or all 1's along its diagonal, and 0's everywhere else (a characteristic
shared by all 3 cases). And life would be a lot easier.

I'm asking WHY T^9=T^8 implies that. You are supposed to prove it. Give a REASON.
 
evilpostingmong said:
Also there are many cases where T is normal, and T9=T8 leads to
T2=T. I mean, T could be an identity, T could be a 0 map,
T could be an orthogonal projection, let me know if I missed something, lol.
what if T7 is 0? So that's out of the question.
I‘m sure you missed a lot, say, a skew-Hermitian.
 
Dick said:
I'm asking WHY T^9=T^8 implies that. You are supposed to prove it. Give a REASON.
Obviously evil thought there are only 3 cases which can be normal.
 
kof9595995 said:
Obviously evil thought there are only 3 cases which can be normal.

Maybe. I just want evilpostingmong to tell me why T^9=T^8 and that the matrix is diagonal tells me the diagonal elements are 0 or 1. That's all. Then I can die happy.
 
kof9595995 said:
I‘m sure you missed a lot, say, a skew-Hermitian.
My bad, I thought he just wanted a normal matrix.
 
evilpostingmong said:
Ok Let T8v=w. Now if T9=T8,
then T9v=T8v. Now we have T9v
=TT8v=Tw. Tw=w. Now if Tw=w, then TTw=Tw.

That is horrible beyond belief. You really don't listen to yourself talk, do you? Anybody that would believe that's a proof of anything, and I'm not even sure what you are trying to prove with that, would be a complete sucker. How many hundreds of posts have we had? You might recall I asked you to prove the entries of a diagonal matrix were 0 or 1. What's that garbage?
 
I mean I know that when you multiply one of the matrices I mentioned earlier
by itself (no matter how many times) you'd get the same matrix. I just don't know how to prove it.
 
evilpostingmong said:
I mean I know that when you multiply one of the matrices I mentioned earlier
by itself (no matter how many times) you'd get the same matrix. I just don't know how to prove it.

If you don't know how to prove it, just say so. Don't post nonsense instead. WHY do you do that! It's REALLY annoying. I would suggest you reread the whole thread, take a stress pill and think about it. Anybody else that want's to chip in here, take over. Actually, I'm the one who should take a stress pill.
 
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evilpostingmong said:
I mean I know that when you multiply one of the matrices I mentioned earlier
by itself (no matter how many times) you'd get the same matrix. I just don't know how to prove it.

If the entries of a diagonal matrix are 1's and 0's would you know how to prove it?
 

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