Matrix Diagonalization ? Suppose A is diagonizable with independent eigenvectors V = [v 1 , . . . , v n ] ? – use similarity transformations to diagonalize dynamics matrix x˙ = Ax x˙ d = A d x d ? ? λ 1 ? ? ? ? ? ? ? ? ? ? ? ? Δ = Λ = A d V ?1 AV = · λ n – Corresponds to change of state from x to x d = V ?1 x ? System response given by e At , look at power series expansion At = V ΛtV ?1 2 (At) 2 = (V ΛtV ?1 )V ΛtV ?1 = V Λ t 2 V ?1 ? (At) n = V Λ n t n V ?1 1 At e = I + At + (At) 2 + . . . 2 ? ? ? ? 1 2 2 V ?1 V I + Λ + Λ t + . . . = ? ? 2 ?? λ 1 t e ? ? ? ? ? ? · λ n t e ? ? ? ? ? ? V e Λt V ?1 V ?1 = V= 1 ? ? ? ? Taking Laplace transform, ? ? 1 s?λ 1 ? ? ? ? ? ? ? ? ? ? ? ? (sI ? A) ?1 = V V ?1 · 1 s?λ n R i n i=1 s ? λ i where the residue R i = v i w i T , and we de?ne = ?? T w 1 ? ? ? ? ? ? ? ? ? ? ? ? , V ?1 = . . V = v 1 . . . v n . T w n ? Note that the w i are the left eigenvectors of A associated with the right eigenvectors v i ???? λ 1 λ 1? ? ? ? ? ? · ? ? ? ? ? ? ? V ?1 A = ? ? ? ? ? ? ? ? ? ? ? ? V ?1 AV = V · λ n λ n ?????? T T λ 1 w w 1 ? ? ? ? ? ? ? ? ? ? ? ? A = ? ? ? ? ? ? · ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 . . . . . . T T λ n w w nn T where w i A = λ i w T i 2 ? ? ? So, if x˙ = Ax, the time domain solution is given by n = e Tλ i t v i w i x(0) dyad x(t) i=1 n x(t) = [ T i x(0)]e λ i t v i w i=1 ? The part of the solution v i e λ i t is called a mode of a system – solution is a weighted sum of the system modes – weights depend on the components of x(0) along w i ? Can now give dynamics interpretation of left and right eigenvectors: Av i = λ i v i , w i A = λ i w i , w i T v j = δ ij so if x(0) = v i , then n ? x(t) = (w T i x(0))e λ i t v i i=1 λ i t = e v i ? so right eigenvectors are initial conditions that result in relatively simple motions x(t). With no external inputs, if the initial condition only disturbs one mode, then the response consists of only that mode for all time. 3 ? ? ? ? ? ? ? ? ? ? ? ? ? If A has complex conjugate eigenvalues, the process is similar but a little more complicated. ? Consider a 2x2 case with A having eigenvalues a ± bi and associated eigenvectors e 1 , e 2 , with e 2 = eˉ 1 . Then ?? a + bi 0 ?1 ? ? ? ? A = e 1 e 2 e 1 e 2 0 a ? bi a + bi 0 ? ? ?1 ≡ TDT ?1 ? ? ? ? = e 1 eˉ 1 e 1 eˉ 1 0 a ? bi Now use the transformation matrix ? ???? ? ? 1 ?i ? ? M ?1 = ? ? 1 1 ? ? M = 0.5 1 i i ?i Then it follows that ? A = TDT ?1 = (TM)(M ?1 DM)(M ?1 T ?1 ) = (TM)(M ?1 DM)(TM) ?1 which has the nice structure: ?? a b ?1 ? ? ? ? A = Re(e 1 ) Im(e 1 ) Re(e 1 ) Im(e 1 ) ?b a where all the matrices are real. ? With complex roots, the diagonalization is to a block diagonal form. 4 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? For this case we have that ? ?? ? ? cos(bt) sin(bt) ?1 At = Re(e 1 ) Im(e 1 ) e at ? ? Re(e 1 ) Im(e 1 )e ? sin(bt) cos(bt) ?1 Note that Re(e 1 ) Re(e 1 ) Im(e 1 ) Im(e 1 ) is the matrix that inverts ? ?? Re(e 1 ) Im(e 1 ) ?1 ? 1 0 ? ? ? ? Re(e 1 ) Im(e 1 ) = 0 1 ? So for an initial condition to excite just this mode, can pick x(0) = [Re(e 1 )], or x(0) = [Im(e 1 )] or a linear combination. ? Example x(0) = [Re(e 1 )] ?? ? ? cos(bt) sin(bt) x(t) = e At x(0) = Re(e 1 ) Im(e 1 ) e at ? ? · ? sin(bt) cos(bt) ?1 Re(e 1 ) Im(e 1 ) [Re(e 1 )] ???? cos(bt) sin(bt) 1 = Re(e 1 ) Im(e 1 ) e at ? ? ? ? ? ? ? ? ? ? sin(bt) cos(bt) cos(bt) 0 ? ? ? ? ? at = e Re(e 1 ) Im(e 1 ) ? sin(bt) at = e (Re(e 1 ) cos(bt) ? Im(e 1 ) sin(bt)) which would ensure that only this mode is excited in the response 5 ? ? Example: Spring Mass System ? Classic example: spring mass system consider simple case ?rst: m i = 1, and k i = 1 M1 k1 k2 k3 k4 k5 M3 M2 Z1 Z3 Z2 x = z 1 z 2 z 3 z˙ 1 z˙ 2 z˙ 3 ?? 0 I ? ? ? ? A M = diag (m i )= ?M ?1 K 0 k 1 + k 2 + k 5 ? ? ?k 5 ?k 2 ? ? ? ? ? ? ? ? ? ? ? ? K ?k 5 k 3 + k 4 + k 5 ?k 3 = ?k 2 ?k 3 k 2 + k 3 ? Eigenvalues and eigenvectors of the undamped system λ 1 = ±0.77i λ 2 = ±1.85i λ 3 = ±2.00i v 1 v 2 v 3 1.00 1.00 1.00 1.00 1.00 ?1.00 1.41 ?1.41 0.00 ±0.77i ±1.85i ±2.00i ±0.77i ±1.85i ?2.00i ±1.08i ?2.61i 0.00 1 - - - ? Initial conditions to excite just the three modes: x i (0) = α 1 Re(v i ) + α 2 Im(v 1 ) ?α j ∈ R – Simulation using α 1 = 1, α 2 = 0 ? Visualization important for correct physical interpretation ? Mode 1 λ 1 = ±0.77i M 1 M 3 M 2 – Lowest frequency mode, all masses move in same direction – Middle mass has higher amplitude motions z 3 , motions all in phase 0 1 2 3 4 5 6 7 8 9 10 ?0.8 ?0.6 ?0.4 ?0.2 0 0.2 0.4 0.6 time displacement V1 z1 z2 z3 2 ? Mode 2 λ 2 = ±1.85i M 1 M 3 M 2 -  - – Middle frequency mode has middle mass moving in opposition to two end masses – Again middle mass has higher amplitude motions z 3 0 1 2 3 4 5 6 7 8 9 10 ?0.4 ?0.3 ?0.2 ?0.1 0 0.1 0.2 0.3 0.4 time displacement V3 z1 z2 z3 3 ? Mode 3 λ 3 = ±2.00i M 1 M 3 M 2 -  0 – Highest frequency mode, has middle mass stationary, and other two masses in opposition 0 1 2 3 4 5 6 7 8 9 10 ?0.4 ?0.3 ?0.2 ?0.1 0 0.1 0.2 0.3 0.4 time displacement V2 z1 z2 z3 ? Eigenvectors with that correspond with more constrained motion of the system are associated with higher frequency eigenvalues 4