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), . . . ∈ Cn be a sequence of vectors which evolve according to the equation v(k + 1) = Av(k) for some fixed A ∈ Mnn(C) and all k ≥ 0. Question Given initial condition v(0) can we find a nice formula for v(k) as a fn

• f k.

First answer As in 1st year, v(k) = Akv(0). Role of Jordan forms The question thus reduces to finding a nice formula for Ak as a fn of k. Now we know there is a Jordan canonical form J = C −1AC for some C ∈ GLn(C). Hence A = CJC −1 and Ak = CJkC −1 so we are reduced to computing a nice formula for Jk.

Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 1 / 9

Powers of direct sums

The following result reduces the computation of Jk, to the case of Jordan blocks.

Prop

For i = 1, . . . , r, let Ti, Si : Vi − → Vi be linear maps. Consider the direct sums T = T1 ⊕ . . . ⊕ Tr, S = S1 ⊕ . . . ⊕ Sr : ⊕iVi − → ⊕iVi.

1

S ◦ T = (S1 ◦ T1) ⊕ . . . ⊕ (Sr ◦ Tr)

2

T k = T k

1 ⊕ . . . ⊕ T k r

• Proof. 1) =

⇒ 2) by induction. To see 1), let (v1, . . . , vr)T ∈ ⊕iVi & just observe (S ◦ T)(v1, . . . , vr)T = S(T1v1, . . . , Trvr)T = ((S1 ◦ T1)v1, . . . , (Sr ◦ Tr)vr)T = ((S1 ◦ T1) ⊕ . . . ⊕ (Sr ◦ Tr))(v1, . . . , vr)T. Ex You can prove this by multiplying block diagonal matrices too. E.g. ( 2 0

0 3 )k = ((2) ⊕ (3))k =

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

Powers of the Jordan block Jn(0)

E.g. To get a feel of what’s going on we compute J3(0)3 =

Prop

Let N = Jn(0) & (bij) = Nk for some k ∈ N. Then Nk has all entries zero except along the j − i = k “diagonal” where we have bij = 1.

• Proof. You can just compute this or note the following more enlightening

argument. N = Jn(0) : Fn − → Fn : (x1, . . . , xn)T → (x2, . . . , xn, 0)T is the “shift co-ords up by 1” linear map. Iterating this k times gives Nk : Fn − → Fn : (x1, . . . , xn)T → (xk+1, . . . , xn, 0, . . . , 0)T is the “shift co-ords up 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 J4(0)

Corollary-Defn

A square matrix N ∈ Mnn(F) is nilpotent if for some k we have Nk = 0. The Jordan block Jn(0) is nilpotent since Jn(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 (A + B)k =

• l≥0

k l

• AlBk−l

holds for commuting matrices A, B ∈ Mnn(F) i.e. where AB = BA as long as we interpret k

l

• = 0 for l > k.

Proof is easily seen from any example. E.g. Note that λI3, J3(0) commute since λI3J3(0) = λJ3(0) = J3(0)λI3. (J3(0) + λI3)k =

Daniel Chan (UNSW) Lecture 25: Discrete time systems Semester 2 2013 5 / 9

Powers of the Jordan block Jn(λ)

Applying the binomial formula to Jn(λ) = N + λIn where N = Jn(0) gives

Prop

Let Jn(λ)k = (bij). Then bij = 0 for i > j whilst on the l = j − i ≥ 0 “diagonal”, the entries are all bij = k

l

• λk−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) = Av(k) where A ∈ M44(R) has Jordan canonical form C −1AC = J = J3(2) ⊕ J1(4). Solve for v(k) if C =     1 1 1 1 −1 1 2 −1 1 −1     , v(0) =     2 1     .

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