Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 3 pptx

A matrix J is in Jordan canonical form if all the elements are zero except for Jordanblocks Jk along the diagonal.J = S−1AS,where S is the matrix of the generalized eigenvectors of A and

Computing Generalized Eigenvectors Let λ be an eigenvalue of multiplicity m Let n be the smallest integersuch that

rank (nullspace ((A − λI)n)) = m

Let Nk denote the number of eigenvalues of rank k These have the value:

Nk = rank nullspace (A − λI)k

− rank nullspace (A − λI)k−1

One can compute the generalized eigenvectors of a matrix by looping through the following three steps until all thethe Nk are zero:

1 Select the largest k for which Nk is positive Find a generalized eigenvector xk of rank k which is linearlyindependent of all the generalized eigenvectors found thus far

2 From xk generate the chain of eigenvectors {x1, x2, , xk} Add this chain to the known generalized tors

eigenvec-3 Decrement each positive Nk by one

Example 15.3.1 Consider the matrix

χ(λ) =

2 1 − λ −1

2 1 − λ −1

−1



(b) One solution of the system of differential equations is

x(1) =





01

We already have a solution of the first equation, we need the generalized eigenvector η Note that η is onlydetermined up to a constant times xi Thus we look for the solution whose second component vanishes tosimplify the algebra.



 e2t

(d) To find a third solution we substutite the form x = xi(t2/2) e2t+ηt e2t+ζ e2t into the differential equation

x0 = Ax2xi(t2/2) e2t+(xi + 2η)t e2t+(η + 2ζ) e2t = Axi(t2/2) e2t+Aηt e2t+Aζ e2t

(A − 2I)xi = 0, (A − 2I)η = xi, (A − 2I)ζ = η

We have already solved the first two equations, we need the generalized eigenvector ζ Note that ζ is onlydetermined up to a constant times xi Thus we look for the solution whose second component vanishes to

simplify the algebra.

Thus there are two eigenvectors.

−3



Two linearly independent solutions of the differential equation are

x(1) =





102

xi = c1xi1 + c2xi2Now we find the generalized eigenvector, η Note that η is only determined up to a linear combination of

xi1 and xi2 Thus we can take the first two components of η to be zero

We see that we must take c1 = c2 in order to obtain a solution We choose c1 = c2 = 2 A third linearlyindependent solution of the differential equation is

x(3) =





24

...

2 Use the results from part (a) to construct eAt and therefore the solution to the initial value problem above

3 Use the results of part (a) to find the general solution... the form x = xit e2t+η e2t, and find appropriate vectors xi and η This gives

a solution of the system (15.2) which is independent of the one obtained in part. .. λ)2(4 − λ) + + + 3( 1 − λ) − 2(4 − λ) + 2(1 − λ)

= −(λ − 2)3< /sup>