Let $U$ be an $n \times n$ orthogonal matrix. Prove that for any $n \times n$ matrix $A$, the matrices $A_{m} = \frac{1}{m + 1} \sum_{j=0}^{m} U^{-j} A U^{j}$ converge entrywise as $m \to \infty$.
#Convergence of Matrix Series Involving Orthogonal Matrices
jade trout
copper gulchBOT
aryan_1201
wanton summitBOT
Let $U$ be an $n \times n$ orthogonal matrix. Prove that for any $n \times n$ ma...
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high shell
**aryan\_1201**
Hmm well you can use the induite norm which is actually an algebraic norm so basically like you said |||A|||=sup||Ax|| for x in Mn,1(R) ||x||=1 here using the equivalence of norms you can use the Euclidean norm so using this for any m |||Am|||<=1/(m+1) * sum( for j going from 0 to m) of |||U||| * |||U^-1||| ^j * |||A||| using the euclidien norm ||Ux||=||x|| (U being orthogonal it represents a vectorial isométrie in a orthonormal basis) so |||U|||=sup||x||=1 so |||U|||*|||U^-1|||=1 so you just have when summing |||Am|||<=|||A||| so it converges
Damn the notations are badly written
My bad
hollow gate
Pro tip: use a backslash
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|||A|||
Also I am not entirely convinced that the fact that the norm of A_m converging to the norm of A implies that A_m converges to A
Or at least an argument is required
high shell
No I didn’t say that it converges to A but yes I made a mistake here I only proved that the sequence is bounded so we can apply balzano weierstrauss theorem for each coordinate of Am
Also idk if we can apply cesaros theorem or not here I only know it in real analysis but using summations of comparison relations it may work in any vectorial space ? If we do it for a scalar
Im not sure though
I think it does work if un=o(an) with (un) a sequence of vectors and (an) a sequence of positive scalars then depending on the convergence of divergence of the series of an you have Σun=o(Σan) with Σ an being either the rest or partial sum of an
hollow gate
I am not smart enough for this but wouldn't it be a good idea to investigate this convergent subsequence regardless? It seems to me like a productive endeavor given that the conclusion is that A_m converges, so it will converge to the same limit.
high shell
Yes that would be good but the problem is that we don’t know if Am converges I made a mistake we only know that Am is bounded but like I said we can apply cesaros theorem I think
The convergence of Am is equivalent to that of Bm=U^-m * A * U^m
I may be overthinking it
hollow gate
Yes that would be good but the problem is that we don’t know if Am converges I m...
I tried to write down the computations but did not find any conclusive argument to show that A_m - A_phi(m) converges to zero
high shell
Nope I don’t think the convergence is guaranteed in that case if we are in dimension 2 take U in SO2(R) so U is a rotation matrice of angle t which is non zero then U^-m is a rotation of angle -mt and U^m of angle mt well you can see that the sequences cos(mt) and sin(mt) dont converge so you can take a certain matrice A such that U^-m *A * U^m doesn’t converge
hollow gate
Yeah I was about to say the same about your cesaro argument
The nature of the sequence is still not determined
high shell
You can say the same generally an orthogonal matrice is always diagonalisable, and the specter of an orthogonal matrice is on the unit circle so for U in O(n) Sp(U) C {z in C |z|=1} so you can see that U is similar to D=diag(e^it1,….,e^itn) with (ti)i real numbers so U^-m * A *U^m what I want to say is that the powers of U are similar to the powers of D which don’t converge for m—>infinity and via continuity of the matrice product you can see that convergence isn’t guaranteed
hollow gate
You can say the same generally an orthogonal matrice is always diagonalisable, a...
Good point
What about this? Express A_{n+k} as a function of A_n and some terms that go to 0
And then replace with n = phi(m) where phi is an extraction for a converging subsequence of (A_m)
If for any k, (A_{phi(m) + k}) converges to the same limit, then we are arguably closer to a solution already
jade trout
Thanks, everyone, for your help! I've managed to solve the problem now.
high shell
Thanks, everyone, for your help! I've managed to solve the problem now.
What did you find ?
jade trout
Cya