Thus, for a pth-order difference equation, the dynamic multiplier is given by fi re fi denotes the(, 1)element of F3 Example The(1, 1)elements of F is 1 and the(1, 1)elements of F2([91, 2,op 1, 1,0,0) is 1+o2. Thus D+1=o aY+2 01+2 in a pth-order syster For larger values of j, an easy way to obtain a numerical value for the dynamic multiplier aYt+i/awt in terms the eigenvalues of the matrix F. Recall that the eigenvalues of a matrix F are those numbers a for which F-A=0 (1 For example, for p=2 the eigenvalues are the solutions to 入0 10 0入 (1-入)2 2-02)-02=0 on For a general pth-order system, the determinant in(12)is a pth-order ploy- minal in A whose p solutions characterize the p eigenvalues of F. This polyno- mial turns out to take a very similar form to(13 The eigenvalues of the matrix F defines in equation(12)are the values of A that satisfy -01-1-02-2- 中pThus, for a pth-order difference equation, the dynamic multiplier is given by ∂Yt+j ∂Wt = f j 11, where f j 11 denotes the (1, 1) element of F j . Example: The (1, 1) elements of F 1 is φ1 and the (1, 1) elements of F 2 (= [φ1, φ2, ..., φp][φ1, 1, 0, ..., 0]0 ) is φ 2 1 + φ2. Thus, ∂Yt+1 ∂Wt = φ1; and ∂Yt+2 ∂Wt = φ 2 1 + φ2 in a pth-order system. For larger values of j, an easy way to obtain a numerical value for the dynamic multiplier ∂Yt+j/∂Wt in terms the eigenvalues of the matrix F. Recall that the eigenvalues of a matrix F are those numbers λ for which |F − λIp| = 0. (12) For example, for p = 2 the eigenvalues are the solutions to      φ1 φ2 1 0  −  λ 0 0 λ     = 0 or      (φ1 − λ) φ2 1 −λ     = λ 2 − φ2λ − φ2 = 0. (13) For a general pth-order system, the determinant in (12) is a pth-order ploy￾nominal in λ whose p solutions characterize the p eigenvalues of F. This polyno￾mial turns out to take a very similar form to (13). Proposition: The eigenvalues of the matrix F defines in equation (12) are the values of λ that satisfy λ p − φ1λ p−1 − φ2λ p−2 − ... − φp−1λ − φp = 0. 6