Normal operators and self adjointness

[evilpostingmong]

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.

 

[Dick]

The zero matrix is diagonalizable. It's not invertible. And it's not the only such matrix.

 

[evilpostingmong]

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.

 

[kof9595995]

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.

 

[Dick]

i) What is 'reduced' supposed to mean? ii) If A and B are matrices, AB=0 and A is not zero, that doesn't imply B=0.

 

[evilpostingmong]

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.

 

[HallsofIvy]

The definition of "normal" is that TT^T= T^TT. 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.

 

[evilpostingmong]

The definition of "normal" is that TT^T= T^TT. 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.

 

[Hurkyl]

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"?)

 

[Hurkyl]

The definition of "normal" is that TT^T= T^TT. 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)

 

[evilpostingmong]

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.

 

[kof9595995]

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.

 

[evilpostingmong]

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]

Diagonal form (ie if singular, all 0's).

That's not correct...

 

[evilpostingmong]

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.

 

[kof9595995]

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.

 

[Dick]

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?

 

[evilpostingmong]

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.

 

[Dick]

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.

 

[evilpostingmong]

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.

 

[Dick]

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.

 

[kof9595995]

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]

I‘m sure you missed a lot, say, a skew-Hermitian.

Quoi?

 

[kof9595995]

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.

 

[Dick]

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]

I‘m sure you missed a lot, say, a skew-Hermitian.

My bad, I thought he just wanted a normal matrix.

 

[Dick]

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?

 

[evilpostingmong]

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.

 

[Dick]

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.

 

[Dick]

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?

 

[evilpostingmong]

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

I feel dumb. I mean this is such a simple concept...Take T and let it be diagonal. Now if T is diagonal and T^9=T^8, then T^9=/=T^8 if any of T's entries were >1 or <0.

Wait, are you implying that we know that T's entries are 0 's or 1's?
I mean if so: If T^9=T^8 , and we know that T's entries are either 0 or 1,
then T^n=T.

 

[Dick]

If T is diagonal and satisfies T^9=T^8 then a diagonal element t_kk satisfies (t_kk)^9=(t_kk)^8. The equation x^9=x^8 in real numbers has only the solutions x=0 and x=1. BTW kof9595995 brings up a point I've been glossing over. If D is our diagonal matrix, all we really know about the original matrix T is that it is related to D by T=U*D*U^(-1), where U is a unitary transformation. We've pretty much shown D^2=D and (D*)=D. Does that imply T has the same properties?

 

[evilpostingmong]

If T is diagonal and satisfies T^9=T^8 then a diagonal element t_kk satisfies (t_kk)^9=(t_kk)^8. The equation x^9=x^8 in real numbers has only the solutions x=0 and x=1. BTW kof9595995 brings up a point I've been glossing over. If D is our diagonal matrix, all we really know about the original matrix T is that it is related to D by T=U*D*U^(-1), where U is a unitary transformation. We've pretty much shown D^2=D and (D*)=D. Does that imply T has the same properties?

Never knew what a unitary transformation is, so I looked it up on wikipedia.
But yes x^9=x^8 equate when x=1 or 0. I was edging on that two posts ago by
saying that T^9=T^8 when their elements are not less than
0 or not greater than 1. But I guess it was on the shallow side. So about T=U*D*U^-1.
Does D* get multiplied by U* followed by U^-1 cancelling out the "effects" of U? If so, then
all we really knew from the beginning is that D is diagonal, obviously a diagonal matrix doesn't
need to have zeroes or ones on its diagonal.

 

[Dick]

I mean T=UDU^(-1). No *'s. If D^2=D, is it true T^2=T? The real equation x^9=x^8 has only solutions x=0 and x=1. Can you show this? The matrix equation T^9=T^8 has lots of nonzero solutions.

 

[evilpostingmong]

wait a minute, I was thrown off (not your fault) when you told me that we've
pretty much shown D^2=D. I was thinking that the proof was done already.
I was wrong.