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cw544259_handout0

cw544259_handout0 - MS&E 201 Dynamic Systems Problem...

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MS&E 201 Handout #0, Page of 1 of 6 Dynamic Systems January 11, 2001 Problem Session Notes Professor Edison Tse A Review of Linear Algebra Analysis of Dynamic Systems relies upon familiarity with a number of concepts from Linear Algebra and Differential/Difference Equations. This handout is a discussion of those Linear Algebra concepts which are important in MS&E 201. Linear Algebra Matrix algebra is, among other things, a convenient way to represent systems of linear equations - the form in which many dynamic systems problems are framed. Understanding matrix properties is therefore important to the analysis of dynamic systems. To begin with, it is useful to be able to simplify the operation of raising a square matrix to an arbitrary power: A k where A is a square matrix and k is a nonnegative integer. Definition. A diagonal matrix is a square matrix for which a ij = 0 for all i 6 = j . Notice that it is a simple thing to evaluate Λ k , where Λ is a diagonal matrix: the i - th diagonal element of Λ k is simply λ k i where λ i is the i - th diagonal element of Λ. Therefore it would be useful to be able to diagonalize a matrix, if possible, for ease of computation of exponents of the matrix (and it provides insight as well, which is seen when one considers the general solution to a dynamic system). In particular finding (invertible) M and Λ for a given A such that A = M Λ M - 1 (1) would be desirable because A k can then be expressed as A k = ( M Λ M - 1 )( M Λ M - 1 ) . . . ( M Λ M - 1 ) = M Λ( M - 1 M )Λ( M - 1 M ) . . . Λ M - 1 = M Λ k M - 1 . (2) For (2) to apply we must have AM = M Λ. Rewriting M in terms of its column vectors v i we obtain A v 1 v 2 . . . v n = v 1 v 2 . . . v n Λ The above can be rewritten as Av 1 Av 2 . . . Av n = λ 1 v 1 λ 2 v 2 . . . λ n v n So that the M and Λ that we are looking for are described by Av i = λ i v i for all i ( A - λ i I ) v i = 0 for all i. (3)
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MS&E 201 Handout #0, Page of 2 of 6 Dynamic Systems January 11, 2001 Problem Session Notes Professor Edison Tse If ( A - λ i I ) is an invertible matrix then v i = 0. But this cannot be the case because v i is a column vector of M , a nonsingular (invertible) matrix. Therefore ( A - λ i I ) cannot be invertible and we now have an expression for determining λ i : det( A - λ i I ) = 0 for all i. (4) Definition. The eigenvalues of a square matrix A are the set of λ ’s which satisfy (4). Associated with each of these eigenvalues is an eigenvector which together satisfy (3). An eigenvalue and its associated eigenvector together comprise an eigenpair. M is called the modal matrix of A
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