DIAGONALIZABLE LINEAR SYSTEMS AND STABILITY. 1. Algebraic facts. We first recall two descriptions of matrix multiplication. Let A be n n, P be n r, given by its columns: P = [v 1 v 2... v r ], where the v i R n. Then the n r matrix AP, given by columns, is: AP = [Av 1 Av 2... Av r ], Av i R n. We used the fact that A acts as a linear operator on R n (via left multiplication of column vectors). So we can say left multiplication by A corresponds to letting A act on each column vector. Now let P = [v 1... v r ] be n r (given by its column vectors v i R n ), and let B be r r. Then the n r matrix P B is obtained in the following way: the r th column of P B (a vector in R n ) is the linear combination of the v i, with coefficients given by the entries in the r th column of B: column i (B) = (c 1, c 2,..., c r ) column i (P B) = c 1 v 1 +c 2 v 2 +... c r v r R n. Briefly, right multiplication by a square matrix takes linear combinations of columns. Direct sums of subspaces. Let E 1, E 2,..., E r be subspaces R n, with the property that any two intersect only at the zero vector: E i E j = {0}, i j. The subspace of R n consisting of all linear combinations: c 1 v 1 +... + c r v r with c i R, v i E i is called the direct sum of E 1,..., E r, denoted by E 1... E r. In particular, if we have: E 1... E r = R n, and B i is a basis of E i for each i = 1,..., r, the union B = B 1... B r is a basis of R n. 2. Matrices diagonalizable over R. Recall a real number λ is an eigenvalue of the n n matrix A if the vector equation Av = λv has nonzero solutions v R n, v 0. Equivalently the subspace of R n : E(λ) = {v R n Av = λv} is not the trivial subspace {0}. In this case E(λ) is the eigenspace for λ. We say the matrix A is diagonalizable over R if R n admits a basis consisting of eigenvectors of A. Equivalently, if we let E(λ i ), i = 1,..., r be the eigenspaces of A (with λ i the eigenvalues), we have: E(λ 1 )... E(λ r ) = R n. 1
Let B = {e 1,..., e n } be a basis of R n consisting of eigenvectors of A: Ae i = λ i e i, where i = 1,..., n (so the λ i are not necessarily all distinct). Form the n n matrices: P = [e 1 e 2... e n ], Λ = diag(λ 1,..., λ n ). (That is, Λ is the diagonal matrix with the entries given). We have: AP = [Ae 1... Ae n ] = [λ 1 e 1... λ n e n ] = P Λ. (Use both descriptions of matrix multiplication recalled above.) The matrix P is invertible, and this matrix equation has the equivalent forms: A = P ΛP 1, Λ = P 1 AP. This implies, for powers and the exponential of A: A k = P Λ k P 1, e A = P e Λ P 1, where e Λ = diag(e λ 1,..., e λn ). And for any t R: e ta = P e tλ P 1 = P diag(e tλ 1,..., e tλn )P 1. Solution of the ODE system. Any v R n admits an expression as a linear combination: v = c 1 e 1 + c 2 e 2 +... c n e n, c i R, and then the solution of the ODE system x = Ax with initial condition x(0) = v is: x(t) = c 1 e tλ 1 e 1 +... + c n e tλn e n, t R. We use this to give an interpretation of the matrix e ta = [x 1 (t) x 2 (t)... x n (t)] (by columns). With e i as above, let: P = [e 1... e n ], P 1 = [w 1... w n ], w i R n. Since P P 1 = I n (the n n identity matrix), given w r = (w 1 r,..., w n r ) (the r th column vector of P 1 ), we have: w 1 re 1 + w 2 re 2 +... + w n r e n = e 0 r, r = 1,..., n, 2
the r rth vector of the standard basis of R n. Since e tλ is a diagonal matrix, the r th column of the matrix e tλ P 1 is: (e tλ 1 w 1 r, e tλ 2 w 2 r,..., e tλn w n r ). Since e ta = P e tλ P 1, the r th column of e ta is: e tλ 1 w 1 re 1 + e tλ 2 w 2 re 2 +... + e tλn w n r e n. (Just use the second matrix multiplication fact.) Thus we find that the r th column of e ta is the solution x r (t) of the ODE system x = Ax with initial condition x(0) = e 0 r, the r th vector of the standard basis of R n. If v = (v 1, v 2,..., v n ) is the expression of v in the standard basis, by definition this means v = n r=1 vr e 0 r. Then e ta v = n r=1 vr (e ta e 0 r), and e ta e 0 r is (when expressed in the standard basis) the r th column of e ta, which as seen above is x r (t), the solution with initial condition e 0 r. But this implies e ta v = n r=1 vr x r (t), the solution with initial condition n r=1 vr e 0 r, or v. This justifies the claim that e ta v is the solution with initial condition v. Change of coordinates. The invertible matrix P = [e 1... e n ] defines a linear change of coordinates in R n : x(t) = P y(t), y(t) = P 1 x(t). Then: y = P 1 x = P 1 Ax = P 1 AP y = Λy, y(0) = P 1 x(0) = P 1 v. Thus the change of coodinates takes the original ODE system to a diagonal system (decoupled, hence easily solved.) Stable, neutral and unstable subspaces. We define the stable subspace E s as the direct sum of all the eigenspaces E(λ), where λ < 0; the unstable subspace E u as the direct sum of all eigenspaces E(λ) with λ > 0; the neutral subspace E 0 as the eigenspace E(0). From the preceding discussion it is easy to see that: e ta v 0 as t + if v E s ; Thus we have: e ta v as t + if v E u ; e ta v v (constant solution), if v E 0. E s E 0 E u = R n. 3
4. Complex eigenvalues. Let A be an n n matrix with real entries. A complex number µ = a + ib (with b 0) is an eigenvalue of A if there exists a non-zero complex vector v = e + if (e, f R n ) so that Av = µv. The eigenspace of µ is denoted E c (λ). (The subscript c reminds us that these are vectors with complex entries). It follows that the complex conjugate µ is also an eigenvalue, and in fact the spaces E c (µ) and E c ( µ) have the same dimension. We re interested in interpreting this in terms of real numbers and vectors in R n. Considering the real and imaginary parts of the defining equation: we obtain the system: A(e + if) = (a + ib)(e + if), Ae = ae bf, Af = be + af. This may be written in matrix form as: A[e f] = [e f]m, M = [(a, b) (b, a)] (by columns). This implies (exercise) we also have: A k [e f] = [e f]m k for all k 1, e ta [e f] = [e f]e tm. Using the definition of matrix exponential and the Taylor expansions of sin(bt), cos(bt), one shows that: M = [(a, b) (b, a)] e tm = [e ta (cos(bt), sin(bt)) e ta (sin(bt), cos(bt))] (matrices given by their column vectors). This implies: [e f]e tm = [e at (cos(bt)e sin(bt)f) e at (sin(bt)e + cos(bt)f)]. This suggests the solutions of the system x = Ax with initial conditions e and f (vectors in R n, a real eigenpair for µ) are given by (respectively): x e (t) = e at (cos(bt)e sin(bt)f), x f (t) = e at (sin(bt)e + cos(bt)f). Exercise: Verify this directly. In the special case Re(µ) = a = 0 we have: x e (t) = cos(bt)e sin(bt)f, x f (t) = sin(bt)e + cos(bt)f. 4
In general, if µ = a + ib is a complex eigenvalue, we may find a subspace E r (µ) = E r ( µ) ( r for real ) of R n of even dimension 2q, invariant under A, with a basis of the form {e 1, f 1, e 2, f 2,..., e q, f q }, so that for r = 1,..., q we have, as above: Ae r = ae r bf r, Af r = be r + af r. And for the solutions of x = Ax with initial conditions e r, f r (respectively): x er (t) = e at (cos(bt)e r sin(bt)f r ), x fr (t) = e at (sin(bt)e r + cos(bt)f r ). If a = 0, all solutions in E r (µ) = E r (ib) are periodic, with period T = 2π/b (we may assume b > 0). 5. Matrices diagonalizable over C. Definition. An n n matrix A with real entries is diagonalizable over C if R n admits a basis B = {e 1, e 2,..., e p, v 1, w 1, v 2, w 2,..., v q, w q }, where each e i is an eigenvector for a real eigenvalue of A, and each pair v j, w j is a real eigenpair for a complex (non-real) eigenvalue of A. (Of course, p + 2q = n.) Equivalently, the direct sum of eigenspaces E(λ i ) over all real eigenvalues λ i of A and real eigenspaces E r (µ j ) over all complex (non-real) eigenvalues µ j of A is R n : E(λ 1 )... E(λ N ) E r (µ 1 )... E r (µ M ) = R n. If this is the case, letting P = [e 1... e p v 1 w 1... v q w q ] (by columns), P is an invertible n n matrix satisfying: where Λ is the n n matrix: P 1 AP = Λ, A = P ΛP 1, Λ = diag(λ 1, λ 2,..., λ p, M 1, M 2,..., M q ). Here M j is the 2 2 block associated to the complex (non-real) eigenvalue µ j = a j + ib j : M j = [(a j, b j ) (b j, a j )] (by columns). It follows that e ta = P e tλ P 1, where: e tλ = diag(e tλ 1, e tλ 2,..., e tλp, e tm 1,..., e tmq ), 5
with e tm j the 2 2 block: e tm j = e ta j [(cos(b j t), sin(b j t)) (sin(b j t), cos(b j t))] (by columns), j = 1,..., q. We already know how to write down the solutions x(t) of x = Ax for the initial conditions e i, v j, w j, and the solution for general initial conditions follows by taking linear combinations. Stable, unstable and neutral subspaces. We define the stable subsplace E s as the direct sum of all eigenspaces E(λ) for λ R, λ < 0 and E r (µ), for µ complex (non-real) with Re(µ) < 0; the unstable space E u as the direct sum of all eigenspaces E(λ) for λ R, λ > 0 and E r (µ), for µ complex (non-real) with Re(µ) > 0; and the neutral subspace E 0 as the direct sum of the eigenspace E(0) for and the eigenspaces E r (µ), for µ complex (non-real) with Re(µ) = 0. From the earlier discussion, we see that: e ta v 0 as t + if v E s ; e ta v as t + if v E u ; e ta v is bounded for all t R (possibly constant), if v E 0. Thus we also have in the complex- diagonalizable case: E s E 0 E u = R n. Remark. Note that in the case v E 0 we may not conclude the solution x(t) is periodic, as discussed in the next example. 6. An example: two coupled harmonic oscillators. Reference: [Waltman, Section 12.] A simple harmonic oscillator with frequency ω is the mechanical system with one degree of freedom described by the equation: q = ω 2 q, q(t) R, or as a first-order Hamiltonian system: (q, p) = (p, ω 2 q), (q(t), p(t)) R 2. The Hamiltonian is E(q, p) = 1 2 (p2 + ω 2 q 2 ). 6
Consider two such oscillators coupled by a spring with constant k > 0. The second-order equations of motion are: q 1 = ω 2 q 1 k(q 1 q 2 ), q 2 = ω 2 q 2 + k(q 1 q 2 ), (q 1 (t), q 2 (t)) R 2. As a first-order system for x(t) = (q 1, p 1, q 2, p 2 ) R 4 : q 1 = p 1, p 1 = ω 2 q 1 k(q 1 q 2 ), q 2 = p 2, p 2 = ω 2 q 2 + k(q 1 q 2 ). This is also a Hamiltonian system, with Hamiltonian given by: E(q 1, p 1, q 2, p 2 ) = 1 2 [p2 1 + p 2 2 + ω 2 (q 2 1 + q 2 2) + k(q 1 q 2 ) 2 ]. Exercise: Verify this. In matrix form the system can be written as x = Ax, where A is 4 4 and x = (q 1, p 1, q 2, p 2 ) R 4. The eigenvalues of A are ±iω and ±i ω 2 + 2k := ±α, with eigenvectors (1, iω, 1, iω) and (1, iα, 1, iα) (see [Waltman] for the computation). Thus we may take as a basis for E r (iω) R 4 : e 1 = (1, 0, 1, 0), f 1 = (0, ω, 0, ω), and as a basis for E r (ˆω): e 2 = (1, 0, 1, 0), f 2 = (0, α, 0, α). With P = [e 1 f 1 e 2 f 2 ], we have P 1 AP = Λ, where: Λ = diag(m 1, M 2 ), M 1 = [(0, ω) (ω, 0)], M 2 = [(0, α) (α, 0)] (by columns). The solutions in R 4 with initial conditions e 1, f 1, e 2, f 2 are: x e1 (t) = cos(ωt)e 1 sin(ωt)f 1, x f1 (t) = sin(ωt)e 1 + cos(ωt)f 1 ; x e2 (t) = cos(αt)e 2 sin(ˆωt)f 2, x f2 (t) = sin(αt)e 2 + cos(αt)f 2. All other solutions can be obtained from these four by taking linear combinations. Note that each of these solutions is periodic (with least period 2π/ω or 2π/α). But consider the solution x v (t) with initial condition v = (1, 0, 0, 0) = 1 2 (e 1 + e 2 ). Its first component is x 1 v(t) = 1 (cos(ωt) + cos(αt)); 2 7
and this function is not periodic, unless we have a relation of the form: n ω = m, for some natural numbers m, n. α In particular this implies the ratio ω/α is a rational number, which won t be the case for randomly chosen ω and k. For instance, if k = ω 2 (all spring constants equal) we have α = 3ω, so the solution with initial condition corresponding to one body released from rest (p 1 = 0 at q 1 = 1), the other released from rest at its equilibrium position (p 2 = q 2 = 0) is not periodic. All we can say is that x(t) is bounded for all t R. A second conserved quantity. There is a different way to change coordinaets so as to decouple the system, which is physicaly more meaningful. Note that adding the equations for the q i and for the p i we find: (q 1 + q 2 ) = p 1 + p 2, (p 1 + p 2 ) = ω 2 (q 1 + q 2 ). Thus q 1 + q 2 describes a simple harmonic motion with frequency ω. On the other hand, taking the difference of the equations of motion we find: (q 1 q 2 ) = (p 1 p 2 ), (p 1 p 2 ) = (ω 2 + 2k)(q 1 q 2 ). This means q 1 q 2 describes a simple harmonic motion with frequency α = ω 2 + 2k. So we may set: q + = q 1 + q 2, p + = q +, q = q 1 q 2, p = q. This corresponds to making the change of variable x = P y, y = P 1 x, where y = (q +, p +, q, p ) and P 1 is the 4 4 symmetric matrix with orthogonal columns: P 1 = [(1, 0, 1, 0) (0, 1, 0, 1) (1, 0, 1, 0) (0, 1, 0, 1)] (given by columns). The matrix Λ ± = P 1 AP has the decoupled form: Λ ± = diag(m +, M ), M + = [(0, ω 2 ) (1, 0)], M = [(0, α 2 ) (1, 0)] (by columns). It follows that the quantities: E + (q 1, q 2, p 1, p 2 ) = 1 2 [(p 1 + p 2 ) 2 + ω 2 (q 1 + q 2 ) 2 ], E (q 1, q 2, p 1, p 2 ) = 1 2 [(p 1 p 2 ) 2 + α 2 (q 1 q 2 ) 2 ] 8
are conserved by the system. (Exercise: check this directly.) The conserved quantities E, E +, E are not all independent (for instance, E + +E = 2E). But any two of them are. (Exercise: Verify this.) (Note: Two conserved quantities E, F are independent if their gradient vectors E, F are linearly independent at every point.) Geometric interpretation ( advanced ). For each E 0 > 0, the level set {(q 1, p 1, q 2, p 2 ) R 4 E(q 1, p 1, q 2, p 2 ) = E 0 } is a three-dimensional surface in R 4. It is a a bounded surface, since E is a proper function in R 4. It is also regular (well-defined tangent 3-plane at each point), since the gradient E 0 vanishes only at the origin of R 4 (check this), which is not a point of the level set if E 0 > 0. Since E is conserved, any orbit that starts on a particular level set of E stays on that level set. Hence orbits stay on three-dimensional surfaces. Using the coordinates y = (q +, p +, q, p ) in R 4 we can say more. The three-dimensional level sets of E + = 1 2 (p2 + + ω 2 q 2 +) have the form (ellipse) R 2, and similarly for the level sets of E = 1 2 (p2 + α 2 q 2 ). The joint level set: M(c 1, c 2 ) = {y R 4 1 2 (p2 + + ω 2 q 2 +) = c 1, 1 2 (p2 + + ω 2 q 2 ) = c 2 )}, c 1, c 2 > 0 is a two-dimensional surfaces of the type (ellipse) (ellipse), i.e. a torus. The solutions stay on the same torus as the initial condition, and on this torus, the system can be described as the flow along parallel lines of slope α/ω. Exercise (challenging): Generalize this discussion to the case of N oscillators connected by springs, assuming all the spring constants are equal (so ω 2 = k). What are the eigenvalues in this case? (This is a first order Hamiltonian system in R 2N.) 9