Normal operators and self adjointness

[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.

 

[evilpostingmong]

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.

 

[Dick]

No T isn't equal to D. U represents the transformation to the basis where T is represented by the diagonal matrix D.

 

[evilpostingmong]

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?

 

[Dick]

Both basically. U does one, U^(-1) does the other.