Conversation started Feb 19, 2023 at 19:16.
leslie townes
leslie townes
Feb 19, 2023 19:16
sure, yes. with a, b real, the conjugate of a + bi is a - bi. the negative of a + bi is -a - bi. these are equal iff a = -a and -b = -b, i.e., if a is zero.
i should say that i don't quite know what's going on with your T_i and T_j abnove, but i assume at some point you're diagonalizing this thing and getting to something that is the equivalent of comparing diagonal entries.
D.C. the III
D.C. the III
Ok that decomposition makes more sense to me. I was getting lost in what it meant to be an imaginary number
leslie townes
leslie townes
too many darn ways to phrase the spectral theorem.
replace with "zero real part" or equivalently "lambda-bar = -lambda" and you're good to go.
D.C. the III
D.C. the III
The $T_i$ are orthogonol projections, so yea. those are the eignevalues
of T
leslie townes
leslie townes
the less familiar analogue of a complex number being 'real' if it coincides with its conjugate.
that's another way to do the exercise, really. if T satisfies your condition then iT will be self-adjoint, and hence maybe already known to have real eigenvalues. so T has eigenvalues of the form real/i, which are purely imaginary. same characterization, via a different route.
D.C. the III
D.C. the III
Well now that you mention it, the self adjointess was how I first thought of the question. But I thought the negative sign in front of my adjoint nullified self adjoint being true.
Wait...what.....that last sentence has me confused.,,,if the eignevalues are all real due to self-adjointness how are they imaginary?
robjohn
robjohn
Feb 19, 2023 19:24
@Koro: I fleshed out my answer and added it to your question.
leslie townes
leslie townes
an operator that satisfies T = -T* is not generally also going to be self adjoint. but iT will be self-adjoint, and so you can apply what you know about self adjoint operators to iT, then divide everything by i to learn what that says about T.
if k is an eigenvalue of iT, then k/i is eigenvalue of T. if k happens to be real, then k/i will be purely imaginary.
D.C. the III
D.C. the III
All makes sense except for how you got "i" in front of $T$.
leslie townes
leslie townes
no reason except that i wanted to turn something that was skew adjoint into something that was self adjoint. analogous to how if z is a purely imaginary number than iz is a real number.
"multiplication by i" sends the imaginary axis to the real axis.
Koro
Koro
@robjohn thanks a lot :-). I'll check out the details soon.
robjohn
robjohn
@Koro I hope it helps. I have not actually taken a class which covers homotopy.
leslie townes
leslie townes
Feb 19, 2023 19:30
a good analogy to have in mind is that normal operators are kinda like complex numbers (because they diagonalize to things that behave algebraically like tuples of complex numbers, or more generally, complex-valued functions). operators satisfying S* = S are like complex numbers z with z* = z (real numbers), operators satisfying S* = -S are like complex numbers z satisfying z* = -z (purely imaginary numbers), and the algebraic tricks carry over.
e.g. the suspicion that iT will be self adjoint if T is skew adjoint turns out to be true.
D.C. the III
D.C. the III
This is something for me to ponder and internalize. Thanks for the tip and the help
 
Conversation ended Feb 19, 2023 at 19:31.