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Unformatted text preview: 1.9 The Matrix of a Linear Transformation Given a linear transformation T : Rn → Rm, it is possible to ﬁnd a unique matrix A such that T (x) = Ax for all x ∈ Rn Q: How many rows and columns does A have to have? A: m rows, n columns, so A is m × n. Recall e1 = vectors. 1 1 0 0 0 , e2 = 0 , · · · , en = 0 are n × 1 . . . . . . . . . 0 0 1 0 0 Q: Given T , how do we ﬁnd A? A: Very simple – the columns of A are T (e1), T (e2 ), · · · , T (en),
that is, A= T (e1 ) T (e2 )  · · · T (en ) 1 EX: 1 0 0 If T : R3 → R2 sends 0 , 1 , 0 to 0 0 1 −
3 2 1 −2 √ √ 3 2 −1 2 − 23 −1 2 √ , √ 3 2 −1 2 0 , 1 then T (x) = Ax where 0 1 . In particular, √ √ 3 − 2 x1 + 23 x2 1 1 − 2 x1 − 2 x2 + x3 x1 T (x) = A x2 = x3 Another interpretation of this: Recall that 1 0 0 e1 = 0 , e2 = 1 , e3 = 0 are linearly indepen 0 0 1 dent and they span R3, so any x ∈ R3 has the unique representation x = x1 e1 + x2 e2 + x3 e3 Since T is linear, by superposition T (x) = T (x1 e1 + x2 e2 + x3e3 ) = x1 T (e1 ) + x2 T (e2 ) + x3 T (e3 ) 2 = x1 √ − 23 −1 2 +x2 √ 3 2 −1 2 +x3 0 1 = − 3 2 −1 2 √ √ 3 2 1 −2 0 1 x1 x2 = Ax x3 Deﬁnition: The matrix A = T (e1 ) T (e2 )  · · · T (en )
is called the standard matrix for the linear transformation T . EX: Find the standard matrix for the linear transformation T : R2 → R2 that reﬂects points through the line x1 = x2 . Solution: Draw a picture to see what happens to
e1 = 1 0
x2 and e2 = 0 1 :
x2 e2 T (e1 ) e1 x1 T (e2 ) x1 So standard matrix is A =
3 T (e1 ) T (e2 ) = 01 10 , that is, for any x = x1 x2 , T (x) = Ax = 01 10 x1 x2 = x2 x1 EX: Find A for the transformation T : R2 → R2 that rotates vectors an angle θ clockwise. Solution: Draw e1, e2, T (e1 ), T (e2): x2 (0, 1) (sinθ, cosθ ) θ θ (1, 0) x1 (cosθ, −sinθ ) Therefore, the standard matrix is A= T (e1 ) T (e2 ) = cos (θ ) sin (θ ) − sin (θ ) cos (θ ) 4 x2 x2 θ x1 x1 Practice ﬁnding the standard matrix for other transformations. EXs: (i) reﬂection through x−axis: A =
x2 1 0 0 −1
x2 x1 x1 5 (ii) contraction/expansion in x2−direction A = 10 0k x2 k>1 x2 x1 x1 (iii) vertical shear A = 10 k1 x2 k>0 x2 x1 x1 6 Deﬁnition: T : Rn → Rm is
onto if for each z ∈ Rm , there is at least one x ∈ Rn with T (x) = z and onetoone if for each z ∈ Rm , there is at most one (maybe not any) x ∈ Rn with T (x) = z Representing T by the standard matrix A, then T is *onto if one can ﬁnd x such that T (x) = Ax = z , so ** the columns of A span Rm Note: *, ** are equivalent +onetoone if T (x) = Ax = z never has more than one solution, so ++ the columns of A are linearly independent and for z = 0, we see that +++ Ax = 0 has only the solution x = 0 Note: +,++,+++ are equivalent EX: Given T : R2 → R3 deﬁned by T (x) = y where x1 − x2 x1 , y = 2x1 + x2 x= x2 x1 − 3x2
7 (i) Is T onto? (ii) Is T onetoone? Solution:
1−0 1 0−1 T (e1 ) = 2 · 1 + 0 = 2 , T (e2 ) = 2 · 0 + 1 1−3·0 1 0−3·1 1 −1 so A = 2 1 1 −3 (i) A has only 2 pivots (right?) so it cannot solve Ax = z for every z ∈ R3 (look at augmented matrix to see this) so T is not onto. (ii) A has 2 pivots so its columns are linearly independent, so T is onetoone. EX: (1) Is T : R3 → R2 deﬁned by x1 T x2 = x3 √ 3 x1 + 23 x2 2 1 1 − 2 x1 − 2 x2 + x3 √ − a linear transformation? (2) Is it onetoone? 8 (3) Is it onto? Solution: (1) Find T (e1 ) =
T (e3 ) = 0 1 √ − 23 −1 2 so standard matrix is A 3 2 , T (e2 ) = −1 2 √ √ 3 3 −2 0 2 = 1 −1 −2 1 2 √ , T (x) = Ax for every x, so yes, it is a linear transformation (2) Solving the homogeneous problem Ax = 0, −
3 2 1 −2 √ √ 3 2 −1 2 0 1 −→ − √ 3 2 √ 3 2 0 0 −1 1 1 free variable, x3 = t =⇒ x2 = t, x1 = t 1 =⇒ x = t 1 for any t ∈ R, so T is not onetoone. 1 1 −1 For instance, T ( 1 ) = T ( −1 ) = 0 1 −1 (3) A has 2 pivots, so yes, T is onto. Equivalently,√we can solve Ax = b with augmented ma√ b1 − 23 23 0 b1 for any b = trix 1 b2 − 2 − 1 1 b2 2 9 ...
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This note was uploaded on 04/09/2011 for the course MATH 232 taught by Professor Russel during the Spring '10 term at Simon Fraser.
 Spring '10
 Russel
 Linear Algebra, Algebra

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