Chapter 7 Powers of Matrices 7.1 Introduction In chapter 5 we had seen that many natural (repetitive) processes such as the rabbit examples can be described by a discrete linear system (1) ~un+1 = A~un, n = 0, 1, 2,..., and that the solution of such a system has the form n (2) ~un = A ~u0. n In addition, we had learned how to find an explicit expression for A and hence for ~un by using the constituent matrices of A. In this chapter, however, we will be primarily interested in “long range predictions”, i.e. we want to know the behaviour of ~un for n large (or as n → ∞). In view of (2), this is implied by an understanding of the limit matrix n A∞ = lim A , n→∞ provided that this limit exists. Thus we shall: 1) Establish a criterion to guarantee the existence of this limit (cf. Theorem 7.6); 2) Find a quick method to compute this limit when it exists (cf. Theorems 7.7 and 7.8). We then apply these methods to analyze two application problems: 1) The shipping of commodities (cf. section 7.7) 2) Rat mazes (cf. section 7.7) As we had seen in section 5.9, the latter give rise to Markov chains, which we shall study here in more detail (cf. section 7.7). 312 Chapter 7: Powers of Matrices 7.2 Powers of Numbers As was mentioned in the introduction, we want to study here the behaviour of the powers An of a matrix A as n → ∞. However, before considering the general case, it is useful to first investigate the situation for 1 × 1 matrices, i.e. to study the behaviour of the powers an of a number a. (a) Powers of Real Numbers Let us first look at a few examples. n 1 Example 7.1. Determine the behaviour of a for a = 2, − 2 , −1. (a) a = 2 n 1 2 3 4 5 ··· sequence diverges an 2 4 8 16 32 ··· (limit doesn’t exist) 1 (b) a = − 2 n 1 2 3 4 5 ... lim (− 1 )n = 0 n→∞ 2 n 1 1 1 1 1 a − 2 4 − 8 16 − 32 ... converges to 0 (c) a = −1 n 1 2 3 4 5 ··· limit doesn’t exist an −1 1 −1 1 −1 ··· (bounces back and forth) These three examples generalize to yield the following facts. Theorem 7.1. Let a ∈ R. (a) If |a| < 1, then lim an = 0. n→∞ (b) If |a| > 1, then the sequence {an} diverges. (c) If |a| = 1, then lim an exists if and only if a = 1 (and then lim an = 1). n→∞ n→∞ (b) Powers of Complex Numbers Next, we consider the behaviour of the powers cn of a complex number c = a + bi. Here we first need to clarify the notion of the limit of a sequence {cn}n≥1 of complex numbers. Definition. If cn = an + ibn, n = 1, 2,..., is a sequence of complex numbers such that a = lim an and b = lim bn n→∞ n→∞ exist, then we say that c = a + ib is the limit of {cn}n≥1 and write lim cn = a + ib. n→∞ Section 7.2: Powers of Numbers 313 Remarks. 1) If either lim an or lim bn does not exist, then neither does lim cn. n→∞ n→∞ n→∞ 2) The following rule is gives an alternate definition of the limit of the sequence {cn}n≥1 (and is sometimes useful for calculating the limit c): lim cn = c ⇔ lim |cn − c| = 0. n→∞ n→∞ | {z } complex absolute value 2 2 2 [To see this equivalence, write cn = an+ibn and c = a+ib, so |cn−c| = (an−a) +(bn−b) . Then lim cn = c ⇔ lim an = a and lim bn = b ⇔ lim |an − a| = 0 and lim |bn − b| = 0 ⇔ 2 2 2 2 lim(an − a) = 0 and lim(bn − b) = 0 ⇔ lim(an − a) + (bn − b) = 0 ⇔ lim |cn − c| = 0.] 3) We also observe that if c = lim cn exists, then we have n→∞ lim |cn| = |c|. n→∞ 2 2 2 2 2 2 2 2 [For lim |cn| = lim(an + bn) = lim an + lim bn = a + b = |c| , and hence lim |cn| = |c|.] Theorem 7.1 naturally extends to complex numbers in the following way. Theorem 7.2. Let α ∈ C. (a) If |α| < 1 then lim αn = 0. “The powers spiral in to 0.” n→∞ (b) If |α| > 1, then lim αn = ∞. “The powers spiral out to ∞.” n→∞ (c) If |α| = 1, then lim αn does not exist except if α = 1. n→∞ Proof. (a) If |α| < 1, then lim |αn| = lim |α|n = 0, and hence lim αn = 0 by Remark 2) above (with c = 0). (Here we have used the multiplicative rule |αn| = |α|n of the absolute value.) (b) If |α| > 1, then the limit lim |αn| = lim |α|n = ∞ does not exist, and hence neither does the limit lim αn by Remark 3) above. (c) Suppose that β := lim αn exists and that |α| = 1. Then |β| = lim |αn| = lim |α|n = 1, so β 6= 0. Thus we have αβ = α lim αn = lim αn+1 = β, so αβ = β and hence α = 1. Corollary. lim αn exists if any only if α = 1 or |α| < 1. Moreover, n→∞ lim αn = 0 if |α| < 1. n→∞ 5 5 11 11 Example 7.2. Analyze the behaviour of the powers of α = 6 + 12 i, β = 12 + 24 i and γ = cos( 2π ) + i sin( 2π ). 100 100 √ 5 . n Solution. Since |α| = 12 5 = .931694990 < 1, the powers α spiral in to 0, as is shown in figure 7.1. √ 11 . n Since |β| = 24 5 = 1.024864490 > 1, the powers β spiral out to ∞, as is evident in figure 7.1. Finally, since |γ| = 1, the powers γn run around on the unit circle (cf. figure 7.1). 314 Chapter 7: Powers of Matrices 1.5 b b b 1 γ20 bb a a a a a a a a a 2 b a a a a β a a a a b b a 2 a b a a r α a a a a a r r a a a a .5 a β a r α a a a b b a r a β20 a a a r 30 r a 2 b b a r α a γ r 20 r γ a α r a a r r r a −1.5 −a1 −.5 r rr r r .5 1a 1.5 a r r rrr rr a rr rrr rr r a r r r rr r r a b a r r a a r r a a a a r r a b a 10 −.5 a a α r 90a a r r γa a a a a a a b a a a a a a a a a a a a −1 a a b a a a a a a a a a a a a a b β10 b b −1.5 5 11 2π 2π Figure 7.1: Powers of the numbers α = 12 (2+i), β = 24 (2+i) and γ = cos( 100 )+i sin( 100 ) Exercises 7.2. 1. Calculate the following limits whenever they exist: 1 + (−1)ni 5 + (−1)ni (−2)n + i (a) lim ;(b) lim ;(c) lim ; n→∞ 1 + 2ni n→∞ 1 + 210000i n→∞ 1 + 2n n k n ! X 1 + i X (d) lim ;(e) lim eki(1+i) . n→∞ 2 n→∞ k=0 k=0 2. (a) Find the first n for which |(3 + 4i)n| > 107. (b) Find the first n for which |(.3 + .4i)n| < 10−7. 3. Write out a detailed proof of Theorem 7.1. Section 7.3: Sequences of powers of matrices 315 7.3 Sequences of powers of matrices We now extend the results of the previous section about limits of powers of numbers to limits of powers of matrices. Before doing so, we first have to clarify the meaning of a limit for matrices. Definition. Let A1,A2,A3,... be a sequence of m × n matrices. Then we say that A = lim Ak k→∞ (k) if we have A = (aij) and lim aij = aij, for all i, j with 1 ≤ i ≤ m and 1 ≤ j ≤ n, where k→∞ (k) (k) a11 . a1n . Ak = . . (k) (k) am1 . amn 1 ! k 1 0 1 Example 7.3. (a) If Ak = , then lim Ak = because 1 k+1 k→∞ 0 1 k2 k 1 1 k + 1 lim = 0 = lim and lim = 1. k→∞ k k→∞ k2 k→∞ k (−1)k 0 (b) lim does not exist! k→∞ 0 1 Theorem 7.3 (Rules for limits). Suppose that lim Ak = A exists and let B, C be ma- k→∞ trices such that the product CAkB is defined. Then: (a) lim (AkB) = AB, k→∞ (b) lim (CAk) = CA, k→∞ (c) lim (CAkB) = CAB. k→∞ Proof. Use the definition of matrix multiplication and properties of (usual) limits. Definition. A square matrix A is called power convergent if lim Ak exists. k→∞ 1 k 1 k 2 0 0 ( 2 ) 0 0 0 0 0 1 1 k Example 7.4. (a) lim 0 3 0 = lim 0 ( 3 ) 0 = 0 0 0 , so A k→∞ 1 k→∞ 1 k 0 0 4 0 0 ( 4 ) 0 0 0 is power convergent. „ « „ « (b) A = 1 0 is not power convergent, for Ak = 1 0 , and we know that 0 −1 0 (−1)k the sequence (−1)k does not converge.