Re: Orthogonalize[expr]
- To: mathgroup at smc.vnet.net
- Subject: [mg123346] Re: Orthogonalize[expr]
- From: Ray Koopman <koopman at sfu.ca>
- Date: Fri, 2 Dec 2011 07:22:06 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jb7rok$j2b$1@smc.vnet.net>
On Dec 1, 4:25 am, =E9=81=93 =E5=8E=9A <shlwell1... at gmail.com> wrote:
> Question: We diagonalize the matrix
>
> {{L^2*mw^2 + ml^2, a*mw*ml}, {a*mw*ml, r^2*mw^2 + ml^2}}
>
> where L, mw, ml, a and r are real numbers.
>
> As a result, the normalizing orthogonalizion base is
>
> {Cos[the],Sin[the]} and {-Sin[the],Cos[the]}
>
> where Tan[2*the]=2*a*ml/((L^2-r^2)*mw)
>
> But now in mathematica we do it as follows:
>
> In[1]:= Mat = {{L^2*mw^2 + ml^2, a*mw*ml}, {a*mw*ml, r^2*mw^2 + ml^2}}
>
> In[2]:= Orthogonalize[Eigenvectors[Mat]]
>
> The result is quite complicated with Abs[expr] and Conjugate[expr].
> How can we get the desired result?Can you help me?
If your matrix is {{p, r}, {r, q}} then {cos, sin}
for the minimum-absolute-angle rotation is proportional to
{ Sqrt[#^2 + r^2] + Abs@#, If[Negative@#, -r, r] }& [.5(p - q)]
If you don't know the sign of p-q then how do you expect to
avoid testing it and adjusting for it?