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Topic: Constant Norm (Read 482 times) |
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william wu
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Constant Norm
« on: Feb 8th, 2006, 8:44pm » |
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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
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« Last Edit: Feb 9th, 2006, 1:13am by william wu » |
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Eigenray
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Re: Constant Norm
« Reply #1 on: Feb 8th, 2006, 9:50pm » |
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o(n)? (That is, the Lie algebra of O(n), i.e., skew-symmetric matrices.)
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« Last Edit: Feb 8th, 2006, 9:54pm by Eigenray » |
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Eigenray
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Re: Constant Norm
« Reply #4 on: Feb 9th, 2006, 9:14am » |
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To elaborate, we have x(s) = esAx, so we want esA 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). esA is orthonormal when I = (esA)t esA = esA^t esA. Differentiating wrt s, and since At commutes with eA^t, we have esA^t (At+A) esA = 0, and so we find At+A = 0. For sufficiency, run the argument backwards, or just note that eA is clearly orthonormal when At = -A: since A and -A commute, (esA)t esA = e-sAesA = eO = 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.
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william wu
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Re: Constant Norm
« Reply #5 on: Feb 10th, 2006, 2:48am » |
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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( xT x ) ] - Just throw away the square root. Thus we need 0 = (d/dt) [ xT x ] = 2 xT x' = 2 xT A x So 0 = xT A x. - Recall xT A x = xT ((A + AT)/2) x. So when A = -AT, the quadratic form is zero.
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