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Comments:
<0> that's relative
<1> you can make sure that the factorization will involve square roots
<2> Well, if you have an answer to several decimal places, it's rather hard to work it out in your head.
<3> I misread the question from my book
<2> Unless you're some sort of computing machine.
<3> it said that can't be factored /over the set of integers/
<3> so anything involving complex numbers will work as an answer
<1> sure choose (x + a)(x + b) where a and b are not integers, or rationals with the same denominator, and multiply out
<1> not both integers is enough i think, sorry
<0> i think they would want real coefficients
<1> they would be real
<0> so take b= conj(a) and that will do
<1> if you take a, b real
<0> oh you mean a and b real, sure
<1> well, ididn't specfiy
<0> maybe they want *integer* coefficients :)
<1> well, just choose a, b, c such that b^2 - 4ac is not a perfect square, and use ax^2 + bx + c
<1> a, b, c integers
<1> anyone know what the spectral radius of a matrix is ?
<4> http://en.wikipedia.org/wiki/Spectral_radius
<1> wow you could just have explaine
<1> d
<1> you didn't have to write a whole web page
<4> ;)
<4> http://mathworld.wolfram.com/SpectralRadius.html
<4> It looks like the maximum eigenvalue
<1> those were the words i was looking for
<4> Maximum in terms of distance from the origin, I think.
<5> it's not the maximum eigenvalue. it's the radius of the smallest closed disc around 0 that contains the spectrum of the matrix.
<6> That's a great out-of-context sentence.
<0> evilgeek: which is the maximum eigenvalue
<5> Sturgeon: nah.
<0> ***uming 'maximum eigenvalue' means the size of the greater eigenvalue
<5> Sturgeon: [1 1; -1 1] has eigenvalues +/- i.
<0> as Olathe said
<1> what is the spectrum?
<5> Sturgeon: and this stops being true when you move into infinite dimensions.
<0> evilgeek: and its spectrum is 1
<0> spectral radius
<0> i mean
<5> mnvl: the set of all lambda such that M - lambda is not invertible.
<1> there is no such thing as a infinite dimenstional matrix
<5> there is.
<0> evilgeek: sure, it is not true in infinite dimension, but still what they said is correct
<1> show me two peerreviewed references to infinite dimensional matrices
<1> and i will paypal us$1,000 immediately to an account of your choice
<5> pick up any book on operator theory.
<0> mnvl: shhhh
<1> they are not called matrices
<0> stop arguing about trivialities
<1> anyway i was refering to finite dimensional matrices
<5> okay. K. R. Davidson, C^* algebras by example; H. Radjavi and P. Rosenthal, Simultaneous Triangularisation.
<1> ok give me a CC number and i will do a chargeback
<1> paypal is down at the moment
<5> i totally don't have a credit card. just send a cheque to Tor Myklebust, 200 University Ave. W., Waterloo, ON, N2L 1G3.
<1> ok us$1000 is equal to ca$320, rihgt?
<5> you can make your cheque out in US funds.
<4> Ask Google.
<5> hell, you can even deduct 43 cents or whatever for postage.
<0> wtf ca$ is more than us$?
<4> http://www.google.com./search?q=1000+USD+to+CAD
<1> ok in R^n if y is the largest eigenvalue then M - z is invertible for z > y
<1> in infinite dim the not invertible values are still bounded?
<5> nah.
<1> and how DOES one calculate the spectral radius?
<5> the largest eigenvalue thing only works if you're over an algebraically closed field.
<1> yeah but i mean, how do you know the spectral radius isn't infinite?
<1> or can it be
<0> not for finite matrices
<5> use the beurling spectral radius formula: rho(X) = lim_(n->infty) ||X^n||^(1/n)/||X||.
<1> right
<0> mnvl: you can calculate it by limsup ||A^n||^(1/n)
<5> oops, yeah. i didn't mean to divide by ||X||.
<5> note that the limsup there will be equal to the liminf.
<1> that looks just like a supremum, but not necessarily of the spectrum..
<1> let me think
<0> it's not immediately obvious
<5> to prove that the spectral radius is finite, note that, for z > ||X||, (X-z)(1 + X/z + X^2/z^2 + ...) = (1 + X/z + ...)(X-z) = I, or so.
<7> what is "TBA"?
<0> evilgeek: why do you like to introduce complication when he's only working in finite dimension?
<5> rather, |z| > ||X||.
<1> to be announced
<7> ah, alright.
<1> no i originally asked about R^n
<5> Sturgeon: i don't think this is very complicated.
<5> Sturgeon: besides, this works in any banach algebra.
<1> but now i'm curious, i asked about infinite dim
<0> oh ok then.
<1> they only mentioned spectra of operator in the last 3 hour lecture on the last day of the last term of my undergraduate degree
<1> i didn't go
<1> ok, this is not really different to the largest eigenvalue
<0> it is
<5> it totally is. in infinite dimensions, the spectrum includes stuff other than eigenvalues.
<5> consider, for example, the right-shift operator on l^2(N) (the sequences a such that sum_i |a_i|^2 is finite). it has *no* eigenvalues (you should be able to prove this) but it has a nonempty spectrum.
<1> ok, gotcha
<5> oh. i should say what the right-shift is. let e_i be the sequence that has a 1 in the i'th position and zeroes everywhere else. the right-shift sends each e_i to e_{i+1}.
<1> i am looking at
<1> en.wikipedia.org/wiki/Spectrum_of_an_operator
<1> right
<1> this is in the approximate point spectrum
<5> (the spectrum is nonempty because, for instance, 0 is in it.)
<0> huh?
<1> hang on
<0> how is 0 in its spectrum?
<5> Sturgeon: the right-shift is not invertible.
<5> Sturgeon: in particular, it is not surjective.
<0> Oh, i thought you were working on Z
<0> nevermind
<1> because the first term falls off the front
<5> mnvl: right-shift, not left-shift. nothing falls off the front.
<1> oh, i meant if there were an inverse
<5> it's just that the first position is occupied by a zero after hitting anything by the right-shift.
<8> oh my god...
<5> homework problem #1: prove that the spectrum of the right-shift operator is the entire unit disc.
<8> I'm having nightmares...
<1> don't worry genoobie
<1> nightmares aren;t real
<0> homework problem #2: what about the left-shift?
<1> the left shift is not injective
<5> Sturgeon: homework problem #2a: prove that the left-shift and the right-shift are adjoint operators. #2b: prove that the spectrum of the adjoint of an operator T is the conjugate of the spectrum of T. #2c: conclude that the spectrum of the left-shift is equal to the unit disc as well.
<1> conjugate?
<5> you turn a+bi into a-bi.
<1> the **** i thought these were real numbers
<5> hell no.
<1> what kind of perverted version of l^2 is this?
<5> the square-summable complex-valued sequences, of course.
<0> mnvl: even if you have real valued sequences, the spectrum is useful when you allow complex values of "lamda" in the definition
<5> on top of that, the spectrum stops being useful when you start considering its intersection with the real line.
<0> er
<0> i mean when you work with the complex version
<0> well you know what i mean
<1> ok
<1> did someone say something before about algebraic completeness
<5> yes.
<5> linear algebra is a bitch over fields that are not algebraically closed. it follows that operator theory would be infinitely more bitchy over a field that is not algebraically closed.
<1> you don;t need a hilbert space, but you do need a complex banach space... ?
<5> pretty much, yeah.
<5> as usual, lots of extra nice **** happens in the hilbert space case.
<1> well, i was reading a paper about fractal geometry
<1> if they had just used the nice term 'norm of the principal eigenvector'
<1> i would still be ignorant
<1> thanx ppl
<1> evilgeek, my secretary is just writing out the cheque
<9> the sicp lectures are really amazing.
<10> tjcoder: sicp?
<9> structure and interpretation of computer programs
<9> it's kindof a cs cl***ic... mit's computer programming intro book since 1982
<11> hello aperson
<11> Haha.
<3> the equation x^n = 1 has n solutions. Find the 6 solutions to x^6 = 1.
<3> you would use the difference of two squares to factor it, right?
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