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Unformatted text preview: 1.8 Linear Transformations Given m × n matrix A and n × 1 vector x recall that Ax is an m × 1 vector A(x + y ) = Ax + Ay A(cx) = cAx for all x, y ∈ Rn , c ∈ R Consider a function (or transformation, or mapping) T from Rn to Rm , so that x ∈ Rn =⇒ T (x) ∈ Rm Some notation: T (x) is the image of x, the domain of T is all x ∈ Rn for which T (x) is deﬁned, the range of T is all images T (x), T is a linear transformation if T (x + y ) = T (x) + T (y ) and T (cx) = cT (x) for all x, y ∈ Rn , c ∈ R Now, only consider linear transformations deﬁned by T (x) = Ax, for m×n matrix A, which we’ll call matrix transformations. 1 EX: 12 7 0 3 A = −2 5 , u = , b= 4 c = 0 . −1 −5 6 −3 1 12 x1 T (x) = Ax = −2 5 x 2 −5 6 x1 + 2x2 = −2x1 + 5x2 , domain R2 , range R3 −5x1 + 6x2 (i) Find T (u), the image of u (ii) Find x ∈ R2 whose image is b (iii) Is there more than one x whose image is b? (iv) Is c in the range of the transformation T ? Solution:
12 (i) T (u) = Au = −2 5 −5 6 3 −1 1 = −11 −21 (ii) Solve Au = b: Augmented matrix 7 127 12 −2 5 −→ 0 9 18 4 −5 6 −3 0 16 32
2 127 −→ 0 1 2 0 16 32 103 −→ 0 1 2 000 3 2 Back substitution gives x = (iii) No free variables above, so x is unique. (iv) Solve Au = c: Augmented matrix 120 1 20 −2 5 0 −→ 0 9 0 −5 6 1 0 16 1 1 20 120 −→ 0 1 0 −→ 0 1 0 0 16 1 001 ↑ inconsistent No, there is no soln, so c is not in the range of T 3 EXs: Geometric interpretation of various (linear) matrix transformations: T (x) =
x2 20 02 x
x2 2 1 1 x1 2 x1 T (x) = 30 0
1 2 x 4 x2 x2 1
1 2 1 x1 3 x1 T (x) =
x2 12 01 x
x2 shear 1 (2, 1) (3, 1) 1 x1 x1 T (x) = cos (θ ) − sin (θ ) sin (θ ) cos (θ ) x 5 x2 rotation x2 Tu u θ x1 u x1 for θ = 0, T (x) = 10 01 0 −1 1 0 x=x x1 x2 −x 2 x1 for θ = π , T (x) = 2 = 6 x2 x2 x1 x1 rotation counterclockwise by π 2 = 90◦ . Notice that linear transformations preserve vector addition and scalar multiplication: EX:x2
u+v v x2 where is T (u + v )? Tv Tu u x1 x1 Other properties: T (0) = 0
7 T (cu + dv ) = cT (u) + dT (v ) for all c, d ∈ R, u, v . Also, T (c1 v1 + c2 v2 + · · · + cn vn ) = c1 T (v1 ) + c2 T (v2 ) + · · · + cn T (vn ). This is the superposition principle Q: If we know T (v1), T (v2), · · · , T (vn), can we use them
to ﬁnd T (y ) for any y ∈ Span {v1, v2, · · · , vn }? EX: T : R2 → R2 deﬁned by T (x) = rx, r ∈ R is a contraction when 0 ≤ r ≤ 1 and a dilation when r > 0 r=
1 2 contraction 8 r=
x2 1 2 contraction
x2 u T (u) x1 T (w ) w T (v ) x1 v T is a linear transformation because T (cu) = rcu = cT (u) and T (u + v ) = r (u + v ) = ru + rv = T (u) + T (v ) EX: T (x) =
y −1 0 01 x= −1 0 01 x y = −x y
y u T ( u) x x Therefore, reﬂects about y axis 9 EX: T (x) = 10 00 x projects onto xaxis y u T ( u) x EX: Company produces products B and C . materials b1 c1 Unit cost matrix U = b c = b2 c2 = labour b3 c3 overhead costs per dollar of output Producing x1 dollars of product B and x2 dollars of product C =⇒ Total cost T (x) = U x = x1 b + x2 c
10 ...
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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, Transformations

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