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### 09S-60-Lecture06

Course: SG 064747, Fall 2009
School: Cal Poly Pomona
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LECTURE 6: POWERS, ELEMENTARY MATRICES MATH 60 SPRING 2009 1. &lt;a href=&quot;/keyword/linear-transformation/&quot; &gt;linear transformation&lt;/a&gt; s Recall that an m n matrix A implicitly provides us with a function TA : Rn Rm . Indeed, the formula TA (x) = Ax provides us with a rule for transforming vectors in x Rn into certain vectors in Ax Rm . At this point, we would like to be a...

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LECTURE 6: POWERS, ELEMENTARY MATRICES MATH 60 SPRING 2009 1. <a href="/keyword/linear-transformation/" >linear transformation</a> s Recall that an m n matrix A implicitly provides us with a function TA : Rn Rm . Indeed, the formula TA (x) = Ax provides us with a rule for transforming vectors in x Rn into certain vectors in Ax Rm . At this point, we would like to be a bit more formal in our notation and terminology. A function T : Rn Rm is usually referred to as a mapping of a transformation. Although there are many functions that one might study, linear algebra is concerned with a particularly important class of functions: De nition. A function T : Rn Rm is called a <a href="/keyword/linear-transformation/" >linear transformation</a> if T (u + v) = T (u) + T (v) T (cu) = cT (u) (1) for any c R and u, v Rn . A <a href="/keyword/linear-transformation/" >linear transformation</a> T : Rn Rn is called a linear operator on Rn . In particular, note that if A is an m n matrix, then the associated function TA : Rn Rm is a <a href="/keyword/linear-transformation/" >linear transformation</a> . Indeed, note that the rules for matrix arithmetic imply that TA (u + v) = A(u + v) = Au + Av = TA (u) + TA (v) and TA (cu) = A(cu) = cAu = cTA (u). In other words, matrices can encode <a href="/keyword/linear-transformation/" >linear transformation</a> s. Moreover, it turns out that matrix multiplication is another way of representing the composition of <a href="/keyword/linear-transformation/" >linear transformation</a> s. We will see this in a concrete example shortly. Date: Monday 2/2/09. 1 2 MATH 60 SPRING 2009 2. Powers of a Matrix De nition. If A is a square matrix, then we de ne the nonnegative integer powers of A by de ning A0 = I and An = AA A n times for n = 1, 2, 3, . . .. If A is invertible, then we de ne negative integer powers of A by A n = (A 1 )n = A 1 A 1 A 1 . n times Theorems 1.4.7 and 1.4.8 of Anton state that the usual laws of exponents hold for square matrices. However, it is important to observe that we have only de ned An for integer values of n. Therefore expressions like A are not well-de ned (at least at this point in the course). Example 1. Consider the matrix A= 1 0 1 . 1 We will consider the powers of A from both an algebraic and geometric viewpoint. First of all, we note that A is invertible since det A = ad bc =1 1 1 0 =1 = 0. Indeed, the formula for the inverse of a 2 2 matrix tells us that A 1 = 1 0 1 . 1 What is the relationship between A and A 1 in this case? Can we understand these two matrices from a geometric viewpoint? The associated <a href="/keyword/linear-transformation/" >linear transformation</a> s TA : R2 R2 and T(A 1 ) : R2 R2 are given by the formulas TA x y = = and T(A 1 ) x y = = 1 0 1 1 x y 1 0 1 1 x y x+y y x y . y 0 , 1 1 1 Let us apply these transformations to the corners 0 , 0 1 , 0 (2) of the unit square. By studying what happens to these four vectors when we apply A and A 1 (not pictured), we see that the associated <a href="/keyword/linear-transformation/" >linear transformation</a> LECTURE 6: POWERS, ELEMENTARY MATRICES 3 TA : R2 R2 is a shear in the x-direction by a factor of 1 and that T(A 1 ) : R2 R2 is a shear in the x-direction by a factor of 1. In other words, TA shears to the right by a factor of 1 and T(A 1 ) shears to the left by a factor of 1. Geometrically, it is clear that if we compose TA and T(A 1 ) (in either order), we obtain the identity transformation on R2 . This is a geometric way of interpreting the formula AA 1 = A 1 A = I. Indeed, the identity matrix represents the identity transformation on R2 : 1 0 0 1 x y = x . y This geometric picture also hints at what An does. Since TA is a shear in the x-direction by a factor of 1, it stands to reason that T(Ak ) should be a shear in the x-direction by a factor of k. Let us compute A2 , A3 , . . . to check our guess: A2 = = 1 0 1 0 1 0 1 0 1 1 2 , 1 2 1 3 , 1 1 0 k 1 0 0 1 0 k 1 1+k 1 1 0 1 1 1 0 1 1 A3 = A2 A = = and indeed more generally we see that Ak = for k = 1, 2, 3, . . .. Since 1 0 1 0 1 0 1 0 k 1 k 1 k 1 k 1 0 0 1 0 0 1 1 1 = = = = (3) and we see (not pictured) that T(Ak ) is indeed a shear in the x-direction by a factor of k. Similarly, we nd th...

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