Discrete time systems Aim lecture: Show how Jordan canonical forms - PowerPoint PPT Presentation

Discrete time systems Aim lecture: Show how Jordan canonical forms can be useful to study some discrete time systems. Recall from first year the following important example of a discrete system. Example of first order discrete time system Let v (0) , v (1) , . . . ∈ C n be a sequence of vectors which evolve according to the equation v ( k + 1) = A v ( k ) for some fixed A ∈ M nn ( C ) and all k ≥ 0. Question Given initial condition v (0) can we find a nice formula for v ( k ) as a fn of k . First answer As in 1st year, v ( k ) = A k v (0). Role of Jordan forms The question thus reduces to finding a nice formula for A k as a fn of k . Now we know there is a Jordan canonical form J = C − 1 AC for some C ∈ GL n ( C ). Hence A = CJC − 1 and A k = CJ k C − 1 so we are reduced to computing a nice formula for J k . Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 1 / 9

Powers of direct sums The following result reduces the computation of J k , to the case of Jordan blocks. Prop For i = 1 , . . . , r , let T i , S i : V i − → V i be linear maps. Consider the direct sums T = T 1 ⊕ . . . ⊕ T r , S = S 1 ⊕ . . . ⊕ S r : ⊕ i V i − → ⊕ i V i . S ◦ T = ( S 1 ◦ T 1 ) ⊕ . . . ⊕ ( S r ◦ T r ) 1 T k = T k 1 ⊕ . . . ⊕ T k 2 r ⇒ 2) by induction. To see 1), let ( v 1 , . . . , v r ) T ∈ ⊕ i V i & just observe Proof. 1) = ( S ◦ T )( v 1 , . . . , v r ) T = S ( T 1 v 1 , . . . , T r v r ) T = (( S 1 ◦ T 1 ) v 1 , . . . , ( S r ◦ T r ) v r ) T = (( S 1 ◦ T 1 ) ⊕ . . . ⊕ ( S r ◦ T r ))( v 1 , . . . , v r ) T . Ex You can prove this by multiplying block diagonal matrices too. 0 3 ) k = ((2) ⊕ (3)) k = E.g. ( 2 0 Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 2 / 9

Powers of the Jordan block J n (0) E.g. To get a feel of what’s going on we compute J 3 (0) 3 = Prop Let N = J n (0) & ( b ij ) = N k for some k ∈ N . Then N k has all entries zero except along the j − i = k “diagonal” where we have b ij = 1. Proof. You can just compute this or note the following more enlightening argument. → F n : ( x 1 , . . . , x n ) T �→ ( x 2 , . . . , x n , 0) T is the “shift co-ords up N = J n (0) : F n − by 1” linear map. Iterating this k times gives → F n : ( x 1 , . . . , x n ) T �→ ( x k +1 , . . . , x n , 0 , . . . , 0) T is the “shift co-ords up N k : F n − by k ” linear map. One now matches up the matrix representing this lin map with the one in the propn. Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 3 / 9

Nilpotent matrices We compute some powers of J 4 (0) Corollary-Defn A square matrix N ∈ M nn ( F ) is nilpotent if for some k we have N k = 0. The Jordan block J n (0) is nilpotent since J n (0) n = 0. Ex Show that if N is nilpotent, then I − N is invertible. Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 4 / 9

The binomial theorem for commuting matrices Prop The binomial formula � k � ( A + B ) k = � A l B k − l l l ≥ 0 holds for commuting matrices A , B ∈ M nn ( F ) i.e. where AB = BA as long as we � k � interpret = 0 for l > k . l Proof is easily seen from any example. E.g. Note that λ I 3 , J 3 (0) commute since λ I 3 J 3 (0) = λ J 3 (0) = J 3 (0) λ I 3 . ( J 3 (0) + λ I 3 ) k = Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 5 / 9

Powers of the Jordan block J n ( λ ) Applying the binomial formula to J n ( λ ) = N + λ I n where N = J n (0) gives Prop Let J n ( λ ) k = ( b ij ). Then b ij = 0 for i > j whilst on the l = j − i ≥ 0 “diagonal”, � k � λ k − l . the entries are all b ij = l Proof. Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 6 / 9

Example Consider the DTS (discrete time system) v ( k + 1) = A v ( k ) where A ∈ M 44 ( R ) has Jordan canonical form C − 1 AC = J = J 3 (2) ⊕ J 1 (4). Solve for v ( k ) if     1 1 0 1 2 1 − 1 1 2 1     C =  , v (0) =  .     − 1 0 1 0 0   − 1 0 0 0 0 Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 7 / 9

Example continued Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 8 / 9

Example continued Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 9 / 9