By Edwin Bidwell Wilson
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Example text
In chapter 1, we learned that, in the simple one-dimensional case, the solution has the exponential form keλt , k and λ being two scalar constants. 4). 10) where 0 is an m-dimensional null vector. 10) is called an eigenvector of matrix A associated with the eigenvalue λ. 10) has a nontrivial solution u = 0 if and only if det(A − λI) = 0. 12) and P(λ) is called the characteristic polynomial. 4) if λ is an eigenvalue and u is the associated eigenvector of A. 10) which, for given λ, is linear. Notice that for each λ, if u is an eigenvector, so is cu for c = 0.
21) we have Ay = αy − βx. Then defining P = x(1) x(2) y (1) y (2) where x(i) , y (i) denote the ith elements of the vectors x,y, respectively, we 32 Review of linear systems have AP = P α β −β α P −1 AP = α β −β α or . 23) is obviously similar to A (they have the same eigenvalues). In the multi-dimensional case, with real and complex conjugate eigenvalues and eigenvectors (but no multiple eigenvalues), the matrix of coefficients A can be transformed into a similar block diagonal matrix with the general form .
7). 7). 7) if y˙ = A¯ y + c = 0, whence y¯ = −A−1 c. If A is singular, that is, if A has a zero eigenvalue, the inverse matrix A−1 and, therefore, the equilibrium are indeterminate. The discrete-time case is entirely analogous. 8) where yn ∈ Rm , B is a m × m matrix and c ∈ Rm is a vector of constants. We set xn = yn + k so that xn+1 = yn+1 + k = Byn + c + k = Bxn − Bk + c + k. 8). , if none of the eigenvalues of the matrix B is equal to one. 8). 4) x˙ = Ax x ∈ Rm . 4) trivially. This special solution is called the equilibrium solution because if x = 0, x˙ = 0 as well.