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# T - James K Beard

T - James K Beard

Data Fusion II Radar Trackers February 16, 2006is done as follows. First for a two-by-two matrix, the Cayley-Hamilton theorem gives us2A − ( λ1+ λ2)⋅ A+ λ1⋅λ2⋅ I = 0(1.13)or,2A = ( λ1+ λ2)⋅A−λ1⋅λ2⋅ I(1.14)This equation means that the matrix exponential series as given by Equation (2.1-48) onpage 20 can be collapsed to a linear combination of I and A. Since we are looking atexp At ⋅ , we use( )2 2( At) ( λ1 λ2) t( At)λ1 λ2t INote that the eigenvalues of ( At ⋅ ) are the eigenvalues of A multiplied by t.⋅ = + ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ (1.15)1.2.1 General Scalar SolutionThe Cayley-Hamilton theorem tells us that the matrix itself satisfies its characteristicequation, but we know also that both the eigenvalues also satisfy the characteristicequation. Since the characteristic equation is used to collapse the series for theexponential into a polynomial of order N-1 (a first-order polynomial for two by twomatrices), we also haveexp ( λi⋅ t) = a1( t) + a2( t)⋅ λi, i = 1, 2(1.16)We can write this equation down for both eigenvalues and solve for a 1 (t) and a 2 (t):exp( λ1⋅ t) = a1( t) + a2( t)⋅λ1(1.17)exp( λ2⋅ t) = a1( t) + a2( t)⋅λ2is a set of two linear equations in a 1 (t) and a 2 (t), and can be solved for a 1 (t) and a 2 (t).The solution isλ1⋅ exp( λ2⋅t) −λ2⋅exp( λ1⋅t)a1() t =λ1− λ2(1.18)exp( λ1⋅t) −exp( λ2⋅t)a2() t =λ1−λ2These can be substituted into the two linear equations in a 1 (t) and a 2 (t) to verify that theyexp At ⋅ term by term and show thatare correct. We can differentiate the series for ( )dexp( At) Aexp( At)⋅ = ⋅ ⋅ (1.19)dtand use this to verify the solution we have for a 1 (t) and a 2 (t). Note that a 1 (t) and a 2 (t) aswe have found them are the same as given in the problem statement when it gives is “theform a 1 (t) and a 2 (t) (a hint that there may be a sign difference or a proportionality factorbetween the equations given there and the exact solution as we have found here), but thesign of the form for a 1 (t) given in the problem statement is incorrect.1.2.2 General Matrix SolutionWe can extend the result to higher order matrices by simply writing the set of two linearequations in equations in a 1 (t) and a 2 (t) as a set of N linear equations in a set of N timefunctions:Page 3 of 14

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