Normal operators and self adjointness

  • #26
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I‘m sure you missed a lot, say, a skew-Hermitian.
My bad, I thought he just wanted a normal matrix.
 
  • #27
Dick
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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?
 
  • #28
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.
 
  • #29
Dick
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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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  • #30
Dick
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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?
 
  • #31
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.
 
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  • #32
Dick
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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?
 
  • #33
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.
 
  • #34
Dick
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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.
 
  • #35
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.
 
  • #36
But I guess if T=UDU^(-1) and U is unitary, then T=D. So
T^2=(UDU^(-1))^2=D^2. If D^2=D, then T^2=T.
 
  • #37
Dick
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No T isn't equal to D. U represents the transformation to the basis where T is represented by the diagonal matrix D.
 
  • #38
No T isn't equal to D. U represents the transformation to the basis where T is represented by the diagonal matrix D.
Oh so you mean U transforms D into a matrix that represents the basis where D came from, right?
Or is it that U transforms D into a matrix that represents T?
 
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  • #39
Dick
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Both basically. U does one, U^(-1) does the other.
 

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