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        riddles >> putnam exam (pure math) >> Constant Norm
        
(Message started by: william wu on Feb 8^th, 2006, 8:44pm)

Title: Constant Norm
Post by william wu on Feb 8^th, 2006, 8:44pm

Consider the matrix differential equation

x' = A x

where x is a state vector in Rⁿ, x' is the derivative of that vector with respect to time, and A is an n-by-n matrix. Find conditions on A which guarantee that the Euclidean norm || x ||₂ remains constant.

Source: Stephen Boyd, Stanford EE Quals 2006

Title: Re: Constant Norm
Post by Eigenray on Feb 8^th, 2006, 9:50pm

o(n)? (That is, the [hide]Lie algebra of O(n)[/hide], i.e., [hide]skew-symmetric[/hide] matrices.)

Title: Re: Constant Norm
Post by towr on Feb 9^th, 2006, 12:50am

Which norm is that? ||x||₂ (i.e euclidean distance?)

Title: Re: Constant Norm
Post by william wu on Feb 9^th, 2006, 1:21am

Yes sorry, Euclidean norm.

That's right Eigenray. I thought it was a cute result :)

Title: Re: Constant Norm
Post by Eigenray on Feb 9^th, 2006, 9:14am

To elaborate, we have
x(s) = e^sAx,
so we want e^sA to be orthonormal for all s, which is precisely the statement that A is in the Lie algebra o(n) associated to the Lie group O(n).

e^sA is orthonormal when
I = (e^sA)^t e^sA = e^{sA^t} e^sA.
Differentiating wrt s, and since A^t commutes with e^{A^t}, we have
e^{sA^t} (A^t+A) e^sA = 0,
and so we find A^t+A = 0. For sufficiency, run the argument backwards, or just note that e^A is clearly orthonormal when A^t = -A: since A and -A commute,
(e^sA)^t e^sA = e^-sAe^sA = e^O = I.

By the same argument, if we had x in Cⁿ, we'd want A to be in u(n), the Lie algebra of U(n), i.e., skew-Hermitian.

Title: Re: Constant Norm
Post by william wu on Feb 10^th, 2006, 2:48am

Another way:

- Want || x || to be constant, so, the derivative of ||x|| with respect to time should be zero. That is,

0 = (d/dt) [ || x || ] = (d/dt) [ sqrt( x^T x ) ]

- Just throw away the square root. Thus we need

0 = (d/dt) [ x^T x ]
= 2 x^T x'
= 2 x^T A x

So 0 = x^T A x.

- Recall x^T A x = x^T ((A + A^T)/2) x. So when A = -A^T, the quadratic form is zero.