Exponents-Commuting

Exponents-Commuting - 1 ) and has the correct initial...

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Linear Systems and Exponentiation of Commuting Matrices October 10, 2008 Proposition. If S and N are commuting n × n matrices (that is, SN = NS ), then e S + N = e S e N . Proof. The method is to consider a linear differential equation associated to the equality we want to prove and to invoke existence and uniqueness of solutions with given initial conditions (we will prove this in class shortly). The strategy is to consider the differential equation ˙ y = ( S + N ) y . (1) With a given initial condition y 0 . Claim: Both x ( t ) := e t ( S + N ) y 0 and z ( t ) := e tS e tN y 0 satisfy this equation and have the same initial conditions y 0 . If this is the case, then by existence and uniqueness of solutions, we would have x ( t ) = z ( t ), or e t ( S + N ) y 0 = e tS e tN y 0 . If we then set t = 1 in this equality and use the arbitrariness of y 0 , we would obtain the desired result. First of all, it is clear that x ( t ) = e t ( S + N ) y 0 satisfies equation (
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Unformatted text preview: 1 ) and has the correct initial conditions because, in general, the solution to u = Au with initial conditions u is u ( t ) = e tA u . To prove the claim and hence the result, it remains to show that z ( t ) also satises equation ( 1 ). It clearly satises the correct initial conditions. To do this, we dierentiate using the product rule: d dt z ( t ) = d dt ( e tS e tN y ) = Se tS e tN y + e tS Ne tN y (2) Now since S and N commute, we have e tS N = Ne tS . This follows easily using the power series expression for e tS . Thus, equation ( 2 ) becomes d dt z ( t ) = Se tS e tN y + Ne tS e tN y = ( S + N ) e tS e tN y = ( S + N ) z This proves the claim. 1...
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This note was uploaded on 01/04/2012 for the course CDS 140A taught by Professor Marsden during the Fall '09 term at Caltech.

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