Vectors, matrices, eigenvalues and eigenvectors 1
( ) ( ) ( ) Scaling a vector: 0.5V 2 0.5 2 1 = 0.5 = = 1 0.5 1 0.5 ( ) ( ) ( ) ( ) Adding two vectors: V + W 2 1 2 + 1 3 = + = = 1 3 1 + 3 4 ( ) ( ) a b λa λb A scalar times a matrix: λ = c d λc λd ( ) ( ) ( ) a b x y a + x b + y A matrix plus a matrix: + = c d z w c + z d + w ( ) ( ) ( ) a b x y ax + bz ay + bw A matrix times a matrix: = c d z w cx + dz cy + dw 2
Hence the product of a matrix times a vector: ( ) ( ) a b x c d y = ( ) ax + by cx + dy This matrix transforms the vector into another vector: Complex scaling of vector b 0.25b y v= a b ( ) w=tv= ( 0.5 0 0 0.25) v= 0.5a ( 0.25b ) Shearing of vector parallel to x-axis y b v= ( a b ) w=tv= ( 1 0.2 0 1 ) v= ( a+0.2b b ) 0.5a a x a a+0.2b x 3
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A system of linear equations: Can be written as: ( 1 2 2 1 ) ( x y { x 2y = 5 2x + y = 10 ) ( ) 5 = 10 From the first we obtain x = 2y 5, which gives in the second: 2(2y 5)+y = 10 or 4y 10+y = 10 or 5y = 20 i.e. y = 4, and hence x = 3. Finally, for the matrix A = ( ) a b c d define det[a] = ad bc and tr[a] = a + d for the determinant and the trace. 5
Forest succession: Gray Birch Blackgum Red Maple Beech Gray Birch 0.05 0.01 0 0 Blackgum 0.36 0.57 0.14 0.01 Red Maple 0.5 0.25 0.55 0.03 Beech 0.09 0.17 0.31 0.96 For example, the fraction of Red Maple trees after 50 years would be 0.5 the fraction of Gray Birch trees, plus 0.25 the fraction of Blackgum trees, plus 0.55 the fraction of Red Maples, plus 0.03 the fraction of Beech trees. 6
Write table as a matrix: 0.05 0.01 0 0 A = 0.36 0.57 0.14 0.01 0.5 0.25 0.55 0.03 0.09 0.17 0.31 0.96 and define the current state of the forest as a vector, e.g., V 0 = ( 1 0 0 0 ) After 50 years the next state of the forest is defined by: V 50 = AV 0 = ( 0.05 0.36 0.5 0.09 ) which is a forest with 5% Gray Birch, 36% Blackgum, 50% Red Maple, and 9% Beech trees. 7
The next state of the forest is V 100 = AV 50 = ( 0.0061 0.2941 0.3927 0.3071 ) After 100 intervals of 50 years, the state is V 5000 = A 100 V0, where A 100 = 0.005 0.005 0.005 0.005 0.048 0.048 0.048 0.048 0.085 0.085 0.085 0.085 0.866 0.866 0.866 0.866 which is a matrix with identical columns. 8
Now consider an arbitrary vector V = (x y z w), where w = 1 x y z, and notice that A 100 V = 0.005 0.005 0.005 0.005 0.048 0.048 0.048 0.048 0.085 0.085 0.085 0.085 0.866 0.866 0.866 0.866 x y z w = 0.005(x + y + z + w) 0.048(x + y + z + w) 0.085(x + y + z + w) = 0.866(x + y + z + w) (0.005 0.048 0.085 0.866), the succession converges into climax state. This climax vector is an eigenvector of the matrix A! 9
Eigenvalue problem: Av = { ax + by = λx cx + dy = λy ( a b c d ) ( ) x y = λ ( ) x y or { (a λ)x + by = 0 cx + (d λ)y = 0 Multiply first with (d λ), and second with b: { (d λ)[(a λ)x + by] = 0 b[cx + (d λ)y] = 0 Subtract second from first: [(d λ)(a λ) bc]x = 0 Because x 0: (d λ)(a λ) bc = 0 10
Characteristic equation: Since A = λ 2 (a + d)λ + (ad bc) = 0 ( ) a b, this can be written as: c d λ 2 trλ + det = 0 Hence: λ 1,2 = tr ± tr 2 4 det 2 11
Numerical example: Av = ( ) ( ) 1 2 x 2 1 y = λ ( ) x y Characteristic equation: tr[a] = 2 and det[a] = 1 4 = 3 Hence: λ 1,2 = tr ± tr 2 4 det 2 = 2 ± 4 + 12 2 λ 1 = 3 and λ 2 = 1 = 2 ± 4 2 12
Corresponding eigenvectors: { (a λ)x + by = 0 cx + (d λ)y = 0 or { y = λ a b x = λ d c x y First eigenvector λ 1 = 3, a = 1, b = 2, c = 2, d = 1: y = 3 1 2 x = x and x = 3 1 2 y = y hence v 1 = ( ) 1 1 Second eigenvector λ 2 = 1, a = 1, b = 2, c = 2, d = 1: y = 1 1 2 x = x and x = 1 1 y = y hence v 2 2 = ( ) 1 1 We only need one of the two equations! 13
Indeed, general case for eigenvectors: { (a λ)x + by = 0 cx + (d λ)y = 0 First equation delivers: ( ) x y = ( ) b a λ Indeed, second equals zero (delivers characteristic equation): bc + (d λ)(a λ) = 0 ( ) ( ) 2 1 Thus, λ 1 = 3, a = 1, b = 2: v 1 = or v 2 1 = 1 ( ) ( ) 2 1 for λ 2 = 1, a = 1, b = 2: v 2 = or v 1 1 2 = 1 14
Special case, diagonal matrix: A = ( ) a 0 : 0 d λ 1,2 = tr ± tr 2 4 det 2 = (a + d) ± (a + d) 2 4ad 2 λ 1,2 = (a + d) ± a 2 + 2ad + d 2 4ad 2 = (a + d) ± (a d) 2 2 λ 1 = a and λ 2 = d 15
Special case, diagonal matrix: A = ( ) a 0 : 0 d Eigenvectors: { (a λ)x + 0y = 0 0x + (d λ)y = 0 λ 1 = a gives (d a)y = 0 or y = 0, i.e., v 1 = λ 2 = d gives (a d)x = 0 or x = 0, i.e., v 2 = ( ) 1 0 ( ) 0 1 16
Linear differential equations The solution of dx(t)/dt = ax(t) is x(t) = Ce at, where C = x(0). Check this: t Ce at = ace at = ax(t) Now two-dimensional systems: { dx/dt = f(x, y) dy/dt = g(x, y) where x(t) and y(t) are unknown functions of time t, and f and g are functions of x and y. 17
An example: { dx/dt = ax + by dy/dt = cx + dy and { dx/dt = 2x + y dy/dt = x 2y where x and y decay at a rate 2, and are converted into one another at a rate 1. In matrix notation: ( ) dx/dt dy/dt = ( ) ( ) a b x c d y 18
We claim that ( ) dx/dt dy/dt = ( ) ( ) a b x c d y has as a general solution: x(t) = C 1 x 1 e λ 1t + C 2 x 2 e λ 2t y(t) = C 1 y 1 e λ 1t + C 2 y 2 e λ 2t or ( ) x(t) y(t) ( ) ( ) x1 = C 1 e λ1t x2 + C y 2 e λ 2t 1 y 2 where λ 1,2 are eigenvalues and (x i y i ) are the corresponding eigenvectors of the matrix given above. Like x(t) = Ce at, this has only one steady state: (x, y) = (0, 0). 19
( ) ( ) ( ) x(t) x1 Notice that the solutions = C y(t) 1 e y λ1t x2 + C 2 1 y 2 are a linear combination of the growth along the eigenvectors. e λ 2t Since x(t) and y(t) grow when λ 1,2 > 0 we obtain: a stable node when both λ 1,2 < 0 an unstable node when both λ 1,2 > 0 an (unstable) saddle point when λ 1 > 0 and λ 2 < 0 (or vice versa) When λ 1,2 are complex, i.e., λ 1,2 = α ± iβ, we obtain a stable spiral when the real part α < 0 an unstable spiral when the real part α > 0 a neutrally stable center point when the real part α = 0 20
( ) dx/dt Example: dy/dt = ( ) ( ) a b x c d y = ( ) ( ) 2 1 x 1 2 y Since tr = 4 and det = 4 1 = 3 we obtain: λ 1,2 = 4 ± 16 12) 2 so λ 1 = 1 and λ 2 = 3. = 2 ± 1 Hence solutions tend to zero and (x, y) = (0, 0) is a stable node. To find the eigenvector v 1 we write: ( ) ( ) b 1 v 1 = = a λ 1 1 or v 1 ( 1 1 ) 21
For v 2 we write v 2 = ( b a λ 2 ) = ( ) 1 1 In combination this gives ( ) ( ) x(t) 1 = C y(t) 1 1 ( ) e t 1 + C 2 e 3t 1 or x(t) = C 1 e t C 2 e 3t y(t) = C 1 e t + C 2 e 3t The integration constants C 1 and C 2 can be solved from the initial condition: i.e., x(0) = C 1 C 2 and y(0) = C 1 + C 2. 22
Let s check this solution: x(t) = C 1 e t C 2 e 3t y(t) = C 1 e t + C 2 e 3t or dx dt = C 1e t + 3C 2 e 3t dy dt = C 1e t 3C 2 e 3t which should be equal to dx dt = 2x+y = 2(C 1e t C 2 e 3t )+C 1 e t +C 2 e 3t = C 1 e t +3C 2 e 3t dy dt = x 2y = C 1e t C 2 e 3t 2(C 1 e t +C 2 e 3t ) = C 1 e t 3C 2 e 3t 23
Remember that we wrote: f(x) f( x) + x f( x) (x x): f(x) x f( x) f( x) x x
The function f(x, y) = 3x x 2 2xy: x f(x, y) = 3 2x 2y and y f(x, y) = 2x and in the point f(1, 1) = 0: x f(x, y) = 1 and y f(x, y) = 2
Generally f(x, y) f( x, ȳ) + x f (x x) + y f (y ȳ) Or, after defining h x = x x and h y = y ȳ: f(x, y) = f( x + h x, ȳ + h y ) f( x, ȳ) + x f h x + y f h y Example: f(x, y) = 3x x 2 2xy, f(1, 1) = 0, x = 1, y = 2 f(1.25, 1.25) = 3.75 1.5625 3.125 = 0.9375 f(1.25, 1.25) 0 1 0.25 2 0.25 = 0.75 26
Consider { dx/dt = f(x, y) dy/dt = g(x, y) close an equilibrium point at ( x, ȳ), i.e., f( x, ȳ) = g( x, ȳ) = 0 Linear approximation of f(x, y) close to the equilibrium: f(x, y) f( x, ȳ) + x f (x x) + y f (y ȳ) As f( x, ȳ) = 0 we obtain For g(x, y) this yields: f(x, y) x f (x x) + y f (y ȳ) g(x, y) x g (x x) + y g (y ȳ) 27
{ dx/dt = f(x, y) dy/dt = g(x, y) became { dx/dt x f (x x) + y f (y ȳ) dy/dt x g (x x) + y g (y ȳ) Since the partial derivatives are merely the slopes of f(x, y) and g(x, y) at the point ( x, ȳ), they are constants that we can write as a = x f, b = y f, c = x g, d = y g Steady states x and ȳ are also constants, with derivatives zero: dx dt = dx dt d x d(x x) = and dy dt dt dt = dy dt dȳ d(y ȳ) = dt dt Hence { d(x x)/dt = a(x x) + b(y ȳ) d(y ȳ)/dt = c(x x) + d(y ȳ) 28
Changing variables to the distances h x = x x and h y = y ȳ: { dhx /dt = ah x + bh y dh y /dt = ch x + dh y having the solution ( ) hx (t) h y (t) ( ) ( ) x1 = C 1 e λ1t x2 + C y 2 e λ 2t 1 y 2 where λ 1,2 and (x i y i ) are the eigenvalues and corresponding eigenvectors of the Jacobi matrix ( ) ( ) x f J = y f a b = x g y g c d 29
Having the solution ( ) hx (t) h y (t) ( ) ( ) x1 = C 1 e λ1t x2 + C y 2 e λ 2t 1 y 2 The Return time is defined by the dominant eigenvalue: 1 T R = max(λ 1, λ 2 ) when λ 1,2 < 0. 30
Having we know that J = λ 1,2 = tr ± D 2 ( ) x f y f x g y g where = ( ) a b c d D = tr 2 4 det Observing that λ 1 + λ 2 = tr[j] and λ 1 λ 2 = det[j], the latter because 1 4 (tr + D)(tr D) = 1 4 (tr2 D) = 1 4 (tr2 tr 2 + 4 det) = det we can classify steady states by just the trace and determinant of their Jacobi matrix. 31
stable node stable spiral det 6 3 5 4 center saddle non stable spiral 1 D=0 2 non stable node tr λ 1,2 = tr ± D 2 D = tr 2 4 det λ 1 + λ 2 = tr λ 1 λ 2 = det 1. if det < 0 then D > 0: λ 1,2 are real with unequal sign: saddle 2. if det > 0, tr > 0 and D > 0 then λ 1,2 > 0: unstable node. 3. if det > 0, tr < 0 and D > 0 then λ 1,2 < 0: stable node. 4. if det > 0, tr > 0 and D < 0 then λ 1,2 > 0: unstable spiral. 5. if det > 0, tr < 0 and D < 0 then λ 1,2 > 0: stable spiral.
Graphical Jacobian: use the signs only J = x f x g f( x + h, ȳ) h g( x + h, ȳ) h y f y g f( x, ȳ + h) h g( x, ȳ + h) = h α β γ δ with tr[j] = α + δ and det[j] = αδ βγ. If tr < 0 and det > 0 the state will be stable. 33
y y y (x,y+h) (x,y) x (x,y) (x+h,y) x (x,y) x a b c J = x f x g f( x + h, ȳ) h g( x + h, ȳ) h y f y g f( x, ȳ + h) h g( x, ȳ + h) = h + 34
The Graphical Jacobian of the Lotka Volterra model: N a c ( ) α β γ 0 e d R a b tr(j) = α < 0 and det(j) = βγ > 0
Finally: The full Jacobian of the Lotka-Volterra model: dr dt = ar br2 crn dn dt = drn en Nullclines: R = 0, N = a br c and N = 0, R = e d Steady states: (R, N) = (0, 0), (R, N) = (a/b, 0) and (R, N) = ( ) e da eb, d dc 36
Steady states of the Lotka-Volterra model N a c ( ) e d, da eb dc N = a br c e d R Note that da > eb a b 37
Lotka Volterra model: dr dt = ar br2 crn = f(r, N) and dn dt = drn en = g(r, N) New variables h R and h N define the distance to the steady state: dr dt = d( R + h R ) dt = dh R dt and dn dt = d( N + h N ) dt = dh N dt dh R dt dh N dt = f( R+h R, N+h N ) f( R, N)+ R f h R + N f h N ( R, N) ( R, N) = g( R+h R, N+h N ) g( R, N)+ R g h R + N g h N ( R, N) ( R, N) 38
Because f( R, N) = 0 and g( R, N) = 0: dh R = dt R f h R + N f h N ( R, N) ( R, N) dh N = dt R g h R + N g h N ( R, N) ( R, N) ( ) ( ) a 2b R c N c R? β J = d N d R = J = e +γ? The solution of the linear system has the form ( ) ( ) ( ) hr (t) R1 = C h N (t) 1 e λ1t R2 + C N 2 e λ 2t 1 N 2 where λ 1,2 are the eigenvalues of J and (R 1 N 1 ) and (R 2 N 2 ) the corresponding eigenvectors.
For ( R, N) = (0, 0) one finds: J = ( ) a 0 0 e with λ 1 = a > 0 and λ 2 = e < 0, i.e., a saddle point. For ( R, N) = (a/b, 0) one finds: ( a ac ) J = b 0 da eb b The eigenvalues are λ 1 = a < 0 and λ 2 = (da eb)/b. Since da > eb this is also a saddle point. 40
For ( R, N) ( ) = ed, da eb dc J = ( be d ce d da eb c 0 one obtains: ) = ( ) b R c R d N 0 = ( ) α β γ 0 Hence trj = b R < 0 and det J = cd R N > 0 which tells us that the non-trivial steady state is stable. If D = tr 2 4 det = (b R) 2 4cd R < 0 this is a stable spiral point. 41
The graphical Jacobian has the same signs: N a c ( ) α β γ 0 e d R a b tr(j) = α < 0 and det(j) = βγ > 0
Complex numbers Quadratic equation: aλ 2 + bλ + c = 0, with roots λ 1,2 = b ± b 2 4ac 2a = b ± D 2a where D = b 2 4ac What if D < 0? Define: i 2 = 1 or equivalently i = 1 Solve λ 2 = 3 by using i 2 = 1: λ 2 = i 2 3 or λ 1,2 = ±i 3 43
So if D < 0 write: λ 1,2 = b ± i D 2a Solve the equation λ 2 + 2λ + 10 = 0: λ 1,2 = 2 ± 4 4 10 2 = 2 ± 36 2 In other words, λ 1 = 1 + 3i and λ 2 = 1 3i. = 2 ± 6i 2 A complex number z is written as z = α + iβ, where α is called the real part and iβ is called the imaginary part. These two solutions are complex conjugates: z 1 = a + ib and z 2 = a ib 44
Argand diagram: complex number as a vector: The complex plane imaginary part complex number z real part Addition of two complex numbers: adding their real parts, and add their imaginary parts. With z 1 = 3 + 10i and z 2 = 5 + 4i: z 1 + z 2 = (3 + 10i) + ( 5 + 4i) = 3 5 + 10i + 4i = 2 + 14i. 45
Multiplication works like (a + bx)(c + dx): z 1 z 2 = (3 + 10i)( 5 + 4i) = 3( 5) + 3 4i + 10i( 5) + 10i4i = 15 + 12i 50i + 40i 2 = 15 38i 40 = 55 38i. Note: (a + ib)(a ib) = a 2 + b 2 If z = a+ib, its modulus z = a 2 + b 2 (magnitude, length vector). Hence z z = z 2. (Used for division). 46
Mandelbrot set: z i = zi 1 2 + c, where c = a + bi is a point in the Argand diagram, and z 0 = 0. Black points remain bounded, colored points keep growing. The color indicates the number of iterations i = 1, 2,..., n required to reach a size of z n. Start with c = 0.5: 0.5, 0.5 2 + 0.5 = 0.75, 0.75 2 + 0.5,... 47
Linear ODEs { dx/dt = ax + by dy/dt = cx + dy λ with 1,2 = tr ± D 2 and { (a λi )x + by = 0 cx + (d λ i )y = 0 λ 1,2 = tr ± i D 2 or λ 1,2 = α ± iβ v 1 = k = k ( ) ( ) b b = k a λ 1 a (α + iβ) ( ) ( ) b 0 ik = kw a α β 1 ikw 2 where w 1 = ( b a α) and w2 = ( 0 β ) correspond to the real and imaginary parts of the eigenvector.
Similarly v 2 = k ( b a λ 2 ) = k ( b ) a (α iβ) = kw 1 + ikw 2 General solution: ( ) x(t) = C y(t) 1 (w 1 iw 2 )e (α+iβ)t + C 2 (w 1 + iw 1 )e (α iβ)t where the constants k are absorbed into C 1 and C 2. Euler s formula: hence e ix = cos x + i sin x or e ix = cos x i sin x e α+iβ = e α e iβ = e α (cos β + i sin β) 49
Hence from ( ) x(t) y(t) = C 1 (w 1 iw 2 )e (α+iβ)t + C 2 (w 1 + iw 1 )e (α iβ)t we obtain ( ) x(t) y(t) = C 1 (w 1 iw 2 )e αt (cos βt + i sin βt) + C 2 (w 1 + iw 2 )e αt (cos βt i sin βt) = e αt [C 1 (w 1 iw 2 )(cos βt + i sin βt) + C 2 (w 1 + iw 2 )(cos βt i sin βt)]. which dies out whenever α = tr/2 < 0. 50
Initial condition where t = 0, e αt = 1, cos βt = 1 and i sin βt = 0, ( ) x(0) = C y(0) 1 (w 1 iw 2 ) + C 2 (w 1 + iw 2 ) = w 1 (C 1 + C 2 ) + iw 2 (C 2 C 1 ), or x(0) = b(c 1 + C 2 ) and y(0) = (a α)(c 1 + C 2 ) + iβ(c 2 C 1 ) from which we solve the complex pair C 1 and C 2. Note that C 1 + C 2 should be real, whereas C 2 C 1 should be an imaginary number. 51
Lotka-Volterra model 1 1 (a) 0.05 (b) 0.05 (c) N dn/dt 0.5 hr, hn 0 0 x(t), y(t) hy hx 0 0 yx 0 0 0 0.5 1 0 0.5 R R 1-0.05 0 10 20 0 5 10 t 15 t 20-0.05 0 10 20 0 5 10 t 15 t 20 dr dt = ar dn br2 crn, = drn en dt With a = b = c = d = 1, e = 0.5, R = 0.5 and N = 0.5, and h R = 0.05 and h N = 0 52
dr dt = ar br2 crn, dn dt = drn en, ( R, N) = with ( ) ( ) b R c R J = = d N. 0 be d ce d da eb c 0 ( e d da eb ), dc For a = b = c = d = 1 and e = 0.5, R = 0.5 and N = 0.5, and ( ) 0.5 0.5 J = with D = 0.75 0.5 0 implying that λ 1,2 = tr ± i D 2 or λ 1,2 = 0.5 ± i 0.75 2 = 0.25 ± i 0.43. Hence α = 0.25 and β = 0.43, the nontrivial state is stable, has a return time of 1/α = 4, and a wave length proportional to 1/β.
v 1 = ( ) 0.5 0.25 i0.43) and v 2 = ( ) 0.5 0.25 + i0.43). ( x(t) y(t)) = e 0.25t [C 1 v 1 (cos 0.43t + i sin 0.43t) + C 2 v 2 (cos 0.43t i sin 0.43t)] x(t) = e 0.25t 0.5[(C 1 + C 2 ) cos 0.43t + (C 1 C 2 )i sin 0.43t] y(t) =... Using t = 0, e 0.25t = 1, cos 0.43t = 1, x(0) = 0.05, y(0) = 0, and sin 0.43t = 0: x(t) = e 0.25t [0.05 cos 0.433t 0.0289 sin 0.433t] y(t) = e 0.25t 0.0577 sin 0.433t 54