spectral radius and Frobenius norm
in [Math]
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From: mateus.oliveira on 16 Apr 2010 16:08 Let R(M) be the spectral radius of a matrix M and |M| be its Frobenius norm, i.e., the usual euclidean norm if we consider M as a vector. Is there any NxN matrix M such that |M| > N * R(M) ? Can someone come up with a simple example, or provide an argument that such a matrix cannot exist? cordially, mateus
From: mateus.oliveira on 16 Apr 2010 16:12 Sorry, I forgot to mention that the matrix M must be symmetric. thanks, On Apr 16, 10:08 pm, "mateus.oliveira" <mateus.olive...(a)gmail.com> wrote: > Let R(M) be the spectral radius of a matrix M and > |M| be its Frobenius norm, i.e., the usual euclidean norm > if we consider M as a vector. > > Is there any NxN matrix M such that > > |M| > N * R(M) ? > > Can someone come up with a simple example, or provide > an argument that such a matrix cannot exist? > > cordially, > > mateus
From: mateus.oliveira on 16 Apr 2010 16:26 And indeed it would be enough to provide an example such that |M| > sqrt(N) * R(M) ? cordially, mateus On Apr 16, 10:12 pm, "mateus.oliveira" <mateus.olive...(a)gmail.com> wrote: > Sorry, I forgot to mention that the matrix M must be symmetric. > > thanks, > > On Apr 16, 10:08 pm, "mateus.oliveira" <mateus.olive...(a)gmail.com> > wrote: > > > Let R(M) be the spectral radius of a matrix M and > > |M| be its Frobenius norm, i.e., the usual euclidean norm > > if we consider M as a vector. > > > Is there any NxN matrix M such that > > > |M| > N * R(M) ? > > > Can someone come up with a simple example, or provide > > an argument that such a matrix cannot exist? > > > cordially, > > > mateus > >
From: George Jefferson on 16 Apr 2010 19:23 "mateus.oliveira" <mateus.oliveira(a)gmail.com> wrote in message news:fdb485b4-de20-4886-9371-a71fed8edd60(a)c36g2000yqm.googlegroups.com... > And indeed it would be enough to provide an example > such that > > |M| > sqrt(N) * R(M) ? > sqrt(Tr(M*M^H)) > sqrt(N)*lim(sqrt(Tr(M^k*(M^k)^H)))^(1/k)) ==> Tr(M*M^H) > N*lim(Tr(M^k*(M^k)^H)^(1/k)) Let A = M*M^H then we have Tr(A) > N*lim(Tr(A^k)^(1/k)) We can assume A is diagonal without loss of generality, Hence we have Tr(A) > N*lim(||A||_k) Tr(A) > N*Max(A) So clearly Tr(A) <= N*Max(A) Anyways, this should get you started. Gotta run or I would have explained it better. Maybe some mistakes. In any case the proof should be rather direct and there may be one much easier.
From: Robert Israel on 16 Apr 2010 19:37
"mateus.oliveira" <mateus.oliveira(a)gmail.com> writes: > And indeed it would be enough to provide an example > such that > > |M| > sqrt(N) * R(M) ? > > cordially, > > mateus > > On Apr 16, 10:12=A0pm, "mateus.oliveira" <mateus.olive...(a)gmail.com> > wrote: > > Sorry, I forgot to mention that the matrix M must be symmetric. > > > > thanks, > > > > On Apr 16, 10:08=A0pm, "mateus.oliveira" <mateus.olive...(a)gmail.com> > > wrote: > > > > > Let R(M) be the spectral radius of a matrix M and > > > |M| be its Frobenius norm, i.e., the usual euclidean norm > > > if we consider M as a vector. > > > > > Is there any NxN matrix M such that > > > > > |M| > N * R(M) ? > > > > > Can someone come up with a simple example, or provide > > > an argument that such a matrix cannot exist? If you mean a real symmetric example, you're out of luck. By the Spectral Theorem, the operator norm ||M|| of a real symmetric matrix M (using the euclidean norm on R^N) is equal to its spectral radius. Now |M|^2 = sum_j |M e_j|^2 <= N ||M||^2. If you allow complex symmetric matrices, you can take [ 0 i 1 ] [ i 0 0 ] [ 1 0 0 ] which has spectral radius 0. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |