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0.1. Linear Transformations the Following Mostly Comes from My Recitation #11 (With Minor Changes) and So Is Unsuited for the Exam

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0.1. Linear Transformations the Following Mostly Comes from My Recitation #11 (With Minor Changes) and So Is Unsuited for the Exam Suggestions for midterm review #3 The repetitoria are usually not complete; I am merely bringing up the points that many people didn’t know on the recitations. 0.1. Linear transformations The following mostly comes from my recitation #11 (with minor changes) and so is unsuited for the exam. I would begin with repeating stuff, as people seem to be highly confused: 0.1.1. Repetitorium 1. A map T : V ! W between two vector spaces (say, R-vector spaces) is linear if and only if it satisfies the axioms T (0) = 0; T (u + v) = T (u) + T (v) for all u, v 2 V; T (au) = aT (u) for all u 2 V and a 2 R (where the R should be a C if the vector spaces are complex). You can leave out the first axiom (it follows from applying the second axiom to u = 0 and v = 0), but the other two axioms are essential. Examples: 0 1 x1 2 3 x1 • The map T : R ! R sending every to @ x1 − 2x2 A is lin- x2 3x2 u ear. (Indeed, for every u 2 R2 and v 2 R2, we can write u = 1 u2 v u v and v = 1 , and then we have T (u + v) = T 1 + 1 = v2 u2 v2 0 1 0 1 u1 + v1 u1 + v1 u1 + v1 T = @ (u1 + v1) − 2 (u2 + v2) A = @ u1 + v1 − 2u2 − 2v2 A, u2 + v2 3 (u2 + v2) 3u2 + 3v2 0 1 u1 u1 v1 which is the same as T (u) + T (v) = T + T = @ u1 − 2u2 A + u2 v2 3u2 0 1 0 1 v1 u1 + v1 @ v1 − 2v2 A = @ u1 + v1 − 2u2 − 2v2 A. This proves the second axiom. 3v2 3u2 + 3v2 The other axioms are just as easy to check.) 1 2 x1 • The map T : R ! R sending every to x1 + x2 − 1 is not linear. x2 (Indeed, it fails the T (0) = 0 axiom. It also fails the other two axioms, but failing one of them is enough for it to be not linear.) x • The map T : R ! R2 sending every x to is not linear. (Indeed, it x2 fails the second axiom for u = 1 and v = 1 because (1 + 1)2 6= 12 + 12.) 2. If V and W are two vector spaces, and if T : V ! W is a linear map, then the matrix representation of T with respect to a given basis (v1, v2,..., vn) of V and a given basis (w , w ,..., wm) of W is the m × n-matrix MT defined as follows: 1 2 For every j 2 f1, 2, . , ng, expand the vector T vj with respect to the basis (w1, w2,..., wm), say, as follows: T vj = a1,jw1 + a2,jw2 + ··· + am,jwm. Then, MT is the m × n-matrix whose (i, j)-th entry is ai,j. For example, if n = 2 and m = 3, then 0 1 a1,1 a1,2 MT = @ a2,1 a2,2 A , a3,1 a3,2 where T (v1) = a1,1w1 + a2,1w2 + a3,1w3; T (v2) = a1,2w1 + a2,2w2 + a3,2w3. 3. What is this matrix MT good for? First of all, it allows easily expanding T (v) in the basis (w1, w2,..., wm) of W if v is a vector in V whose expansion in 1 the basis (v1, v2,..., vn) of V is known. More importantly, once you know the matrix MT, you can use it as a proxy whenever you want to know something about T. For instance, you want to know the nullspace of T. You know how to compute the nullspace of a matrix, but not how to compute the nullspace of a linear map. So you find the nullspace of MT; it consists of vectors of the 0 1 g1 B g2 C B C T form B . C. Then, the nullspace of consists of the corresponding vectors @ . A gn g1v1 + g2v2 + ··· + gnvn. 4. Suppose that V = Rn and W = Rm. These vector spaces already have standard bases, but nothing keeps you from representing a linear map T : Rn ! 1 For instance, if v is one of the basis vectors vj, then the expansion of T vj can be simply read off from the j-th column of MT; otherwise, it is an appropriate combination of columns. 2 Rm with respect to two other bases. There is a formula to represent the map in this case: Let T : Rn ! Rm be a linear map, and let A be the m × n-matrix representation of T with respect to the standard bases (that is, the matrix such that T (v) = Av n n for every v 2 R ). Let (v1, v2,..., vn) be a basis of R , and let (w1, w2,..., wm) m be a basis of R . Let B be the matrix with columns v1, v2,..., vn, and let C be the matrix with columns (w1, w2,..., wm). Then, the matrix representation of T −1 with respect to the bases (v1, v2,..., vn) and (w1, w2,..., wm) is MT = C AB. 0.1.2. Exercises Exercise 0.1. Let f : R2 ! R3 be the linear map which sends every vector 0 1 −1 1 v 2 R3 to @ 2 0 A v. 3 1 2 Consider the following basis (v1, v2) of the vector space R : 1 1 v = , v = . 1 1 2 0 3 Consider the following basis (w1, w2, w3) of the vector space R : 0 1 1 0 0 1 0 0 1 w1 = @ 1 A , w2 = @ 1 A , w3 = @ 0 A . 1 1 1 Find the matrix M f representing f with respect to these two bases (v1, v2) and (w1, w2, w3). 1 1 [Hint: You can use the M = C−1 AB formula here, with B = , T 1 0 0 1 0 0 1 0 1 −1 1 C = @ 1 1 0 A and A = @ 2 0 A. But it is easier to do it directly, and 1 1 1 3 1 the C−1 AB formula does not help in the next exercises.] First solution. We compute the matrix representation directly, using part 2. of the repetitorium above. We have 0 1 −1 1 0 0 1 1 1 f (v ) = f = 2 0 = 2 = 0w + 2w + 2w 1 1 @ A 1 @ A 1 2 3 3 1 4 3 2 and similarly 0 1 1 f (v2) = @ 2 A = 1w1 + 1w2 + 1w3. 3 Thus, 0 0 1 1 M f = @ 2 1 A . 2 1 Second solution. We can apply part 4. of the repetitorium. Here, n = 2, 0 1 −1 1 0 1 0 0 1 1 1 m = 3, T = f , A = 2 0 , B = , and C = 1 1 0 . Thus, @ A 1 0 @ A 3 1 1 1 1 0 0 1 1 −1 M f = C AB = @ 2 1 A. 2 1 For every n 2 N, we let Pn denote the vector space of all polynomials (with real coefficients) of degree ≤ n in one variable x. This vector space has di- mension n + 1, and its simplest basis is 1, x, x2,..., xn. We call this basis the monomial basis of Pn. Exercise 0.2. Which of the following maps are linear? For every one that is, represent it as a matrix with respect to the monomial bases of its domain and its target. (a) The map Ta : P2 ! P2 given by Ta ( f ) = f (x + 1). (b) The map Tb : P2 ! P3 given by Tb ( f ) = x f (x). (c) The map Tc : P2 ! P4 given by Tc ( f ) = f (1) f (x). 2 (d) The map Td : P2 ! P4 given by Td ( f ) = f x + 1 . 1 (e) The map T : P ! P given by T ( f ) = x2 f . e 2 2 e x 0 0 1 2 Finding the expression 0w1 + 2w2 + 0w3 for @ 2 A can be done by solving a system of lin- 4 ear equations (in general, if we want to represent a vector p 2 Rn as a linear combination p = a1w1 + a2w2 + ··· + anwn of n given linearly independent vectors (w1, w2,..., wn), then 0 1 a1 B a2 C B C we can find the coefficients a1, a2,..., an by solving the equation W B . C = p, where W @ . A an denotes the n × n-matrix with columns w , w ,..., w ). But in our case, the vectors w , w , w 1 2 n 0 1 1 2 3 x1 are so simple that we can easily represent an arbitrary vector @ x2 A as their linear combi- x3 0 1 x1 nation, as follows: @ x2 A = x1w1 + (x2 − x1) w2 + (x3 − x2) w3. x3 4 0 (g) The map Tg : P3 ! P2 given by Tg ( f ) = f (x). [There is no part (f) because I want to avoid having a Tf map when f stands for a polynomial.] (h) The map Th : P3 ! P2 given by Th ( f ) = f (x) − f (x − 1). (i) The map Ti : P3 ! P2 given by Ti ( f ) = f (x) − 2 f (x − 1) + f (x − 2). Solution. First, wet us introduce a new notation to clear up some ambiguities. In the exercise, we have been writing f (x + 1) for “ f with x + 1 substituted for x”. This notation could, however, be mistaken for the product of f with x + 1. (This confusion is particularly bad when f is, say, x; normal people would read x (x + 1) as “x · (x + 1)” and not as “x with x + 1 substituted for x”.) So we change this notation to f jx+1.
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