x =. x = x 2 x 1 x 2 x 2

Transcription

1 3 Linear function 31 Introduction In calculus, a vector in the plane R 2 with components 2 and 3 is usually written using notation such as v = 2, 3 For our purposes it turns out to be more convenient to express such a vector as a 2 1 matrix: [ 2 x = 3] More generally, a vector in R n is written as an n 1 matrix When writing vectors in text we usually use the matrix transpose notation to avoid unseemly vertical spacing For instance, we might write x = [6, 1,3,2] T, when we want to say 6 x = The addition and scalar multiplication defined for matrices(section 21) gives an addition and scalar multiplication for vectors, which coincides with the calculus definitions The idea of a function plays a central role in calculus and the same is true for linear algebra For most of the functions in calculus the inputs and outputs are both real numbers, but in linear algebra, the functions we study have inputs and outputs that are vectors For instance, here is a function L from the set R 2 to the set R 3 : () x1 L = x 1 +4x 2 3x x 1 x 2 2 The notation works just like it did in calculus For example, if the input vector is [2,1] T, then the output vector is ([ 2 L = 1]) (2)+4(1) 3(2) (1) = 6 5 (1) 1 x 2 This function satisfies a couple of properties that make it linear, meaning that it is compatible with the addition and scalar multiplication of vectors (the precise definition is given below) Linear functions are the main functions in linear algebra We study them in this section 1

2 3 LINEAR FUNCTION 2 32 Definition and examples Linear function A function L : R n R m is linear if (a) L(x+y) = L(x)+L(y), (b) L(αx) = αl(x), for all x,y R n, α R The notation L : R n R m is used to indicate that the input vectors come from the set R n (= domain of L) and the output vectors are in the set R m (= codomain of L) 321 Example Show that the function L : R 2 R 3 given by L(x) = x 1 +4x 2 3x 1 x 2 is linear Solution First, the input vector x is an element of R 2 (according to the notation L : R 2 R 3 ), so it is of the form x = [x 1,x 2 ] T This is the meaning of x 1 and x 2 in the formula x 2 We need to verify that L satisfies the two properties in the definition of linear function For any x,y R 2, we have () x1 +y L(x+y) = L 1 x 2 +y 2 = (x 1 +y 1 )+4(x 2 +y 2 ) 3(x 1 +y 1 ) (x 2 +y 2 ) (In the formula, x 1 +y 1 plays the role of x 1 and x (x 2 +y 2 ) 2 +y 2 plays the role of x 2 ) = (x 1 +4x 2 )+(y 1 +4y 2 ) (3x 1 x 2 )+(3y 1 y 2 ) (x 2 )+(y 2 ) = x 1 +4x 2 3x 1 x 2 + y 1 +4y 2 3y 1 y 2 x 2 = L(x)+L(y), y 2

3 3 LINEAR FUNCTION 3 so property (a) holds Next, for any x R 2 and α R, we have () αx1 L(αx) = L = (αx 1)+4(αx 2 ) 3(αx αx 1 ) (αx 2 ) 2 (αx 2 ) = α(x 1 +4x 2 ) α(3x 1 x 2 ) = α x 1 +4x 2 3x 1 x 2 α(x 2 ) = αl(x), so property (b) holds Therefore, L is linear 322 Example Show that the function L : R 1 R 2 given by 2x1 L(x) = x 1 is linear Solution For any x,y R 1, we have 2(x1 +y L(x+y) = L([x 1 +y 1 ]) = 1 ) (x 1 +y 1 ) (2x1 )+(2y = 1 ) ( x 1 )+( y 1 ) 2x1 2y1 = + x 1 y 1 = L(x)+L(y), so property (a) holds Next, for any x R 1 and α R, we have 2(αx1 ) L(αx) = L([αx 1 ]) = (αx 1 ) α(2x1 ) 2x1 = = α α( x 1 ) x 1 = αl(x), so property (b) holds Therefore, L is linear x 2 If a is any number, then the function f : R R given by f(x) = ax has as its graph a straight line (through the origin with slope a) In fact, this function is linear in the sense of the above definition (regarding R as the same thing as R 1 ) The next theorem generalizes this statement with the number a being replaced by a matrix A

4 3 LINEAR FUNCTION 4 Theorem Let A be an m n matrix The function L : R n R m defined by L(x) = Ax is linear The function L in the theorem is called the linear function corresponding to the matrix A Proof It should be checked that L makes sense as a function from R n to R m If x is an input vector, then it is an element of R n, and is therefore an n 1 matrix Since A is m n, the product Ax is defined and equals an m 1matrix, which is an element of R m, as desired We now check that L satisfies the two properties of a linear function For any x,y R n, we have L(x+y) = A(x+y) = Ax+Ay = L(x)+L(y), where the second equality is due to the distributive property of matrix multiplication (property (d) in Section 23) This verifies property (a) Next, for any x R n and α R, we have L(αx) = A(αx) = α(ax) = αl(x) where the second equality is due to a property of matrix and scalar multiplication (property (i) in Section 23) This verifies property (b) and finishes the proof that L is linear This gives us another way to check whether a given function is linear: 323 Example Use the last theorem to show that the function L : R 2 R 3 given by L(x) = x 1 +4x 2 3x 1 x 2 is linear Solution We have L(x) = x 1 +4x 2 3x 1 x 2 = where x 2 x 2 A = [ x1 ] = Ax, x 2

5 3 LINEAR FUNCTION 5 Therefore, L is linear by the preceding result The zero vector in R n is the vector 0 = [0,0,,0] T Theorem Let L : R n R m be a function If L is linear, then L(0) = 0 Proof Assume that L is linear We have L(0)+L(0) = L(0+0) = L(0), where the first equality is due to property (a) of a linear function Subtracting L(0) from both sides of this equation gives L(0) = 0, as desired Put another way, the theorem says that if L does not send 0 to 0, then it cannot be linear 324 Example Is the function F : R 1 R 2, given by 2x1 +1 F(x) =, x 1 linear? Explain Solution Note that F(0) = [ [ 2(0) = = 0 (0) 0] 0] (the string says that F(0) 0), so F is not linear according to the preceding theorem 325 Example Is the function F : R 2 R 2, given by x1 x F(x) = 2, x 1 linear? Explain Solution If we can show that the function does not send 0 to 0, then we can quickly conclude that it is not linear (as in the preceding example) However, [ (0)(0) 0 F(0) = = = 0, (0) 0]

6 3 LINEAR FUNCTION 6 so all we know is that F has a chance of being linear We see if we can verify property (a) of a linear function Let x,y R 2 We have () x1 +y F(x+y) = F 1 (x1 +y = 1 )(x 2 +y 2 ) x 2 +y 2 (x 1 +y 1 ) x1 x = 2 +x 1 y 2 +y 1 x 2 +y 1 y 2 x 1 +y 1 We are trying to show that this equals x1 x F(x)+F(y) = 2 x 1 = y1 y + 2 y 1 x1 x 2 +y 1 y 2 x 1 +y 1 Since the first components (in red) do not match up, we suspect that F is not linear We cannot write F(x+y) F(x)+F(y), though, since there are choices for x and y that actually give equality (for instance, x = 0 and y = 0) However, in order to show that F fails property (a) it is enough to give a single counterexample Using inspection, we see that if x 1, x 2, y 1, y 2 are all equal to 1, for instance, then the first components are not equal, so this should give our counterexample Everythingwe havedone up to this point can be consideredscratchwork It was done just to come up with an idea for a counterexample To solve the problem, all we really need to write is this: If x = [1,1] T and y = [1,1] T, then ([ [ [ F(x+y) = F = = 2]) 2] 2] so F is not linear [ [ = F(x)+F(y), 1] 1]

7 3 LINEAR FUNCTION 7 33 Image, Preimage, and Kernel Definition of image Let L : R n R m be a function Let x be a vector in R n The image of x under L is L(x) The image of L (denoted iml) is the set of all images L(x) as x ranges through R n In symbols, iml = {L(x) x R n } In other words, given an input vector x, its image is the corresponding output vector And the image of L is the set of all actual output vectors 331 Example Let L : R 3 R 2 be given by x1 3x L(x) = 2 +2x 3 2x 1 +6x 2 x 3 (a) Find the image of [4,1, 7] T under L (b) Is [ 5,7] T in iml? Explain Solution (a) The image of [4,1, 7] T under L is L 4 1 = 7 (4) 3(1)+2( 7) = 2(4)+6(1) ( 7) 13 5 (b) The question amounts to asking if there is a vector x in R 3 such that

8 3 LINEAR FUNCTION 8 L(x) = [ 5,7] T, that is, x1 3x 2 +2x 3 = 2x 1 +6x 2 x Thisequalityofvectorsholdsifandonlyifthevectors componentsarethe same, so this leads to a system of equations with corresponding augmented matrix , which has row echelon form [ There is no pivot in the augmented column, so a solution x = [x 1,x 2,x 3 ] T exists Therefore, [ 5,7] T is in iml ] Definition of preimage Let L : R n R m be a function Let y be a vector in R m The preimage of y under L (denoted L 1 (y)) is the set of all x in R n that have image under L equal to y In symbols: L 1 (y) = {x R n L(x) = y}

9 3 LINEAR FUNCTION 9 Definition of kernel Let L : R n R m be a function The kernel of L (denoted kerl) is the preimage of 0 under L In other words, kerl is the set of all vectorsin R n that have image under L equal to 0 In symbols: kerl = L 1 (0) = {x R n L(x) = 0} 332 Example Let L : R 3 R 2 be given by x1 3x L(x) = 2 +2x 3 2x 1 +6x 2 x 3 (a) Determine whether the vector [2,0, 3] T is in the preimage of [ 4,8] T under L (b) Find L 1 ([ 5,7] T ) (c) Find ker L Solution (a) Asking whether the vector [2,0, 3] T is in the preimage of [ 4,8] T under L is asking whether L([2,0, 3] T ) = [ 4,8] T Since L (0)+2( 3) 4 4 = =, 2(2)+6(0) ( 3) [2,0, 3] T is not in the preimage of [ 4,8] T (b) We seek the set of all vectors x in R 3 for which L(x) = [ 5,7] T, that is, x1 3x 2 +2x 3 5 = 2x 1 +6x 2 x 2 7 Thisequalityofvectorsholdsifandonlyifthevectors componentsarethe same, so this leads to a system of equations with corresponding augmented

10 3 LINEAR FUNCTION 10 matrix [ which has reduced row echelon form (RREF) The preimage of [ 5,7] T is the solution set of the corresponding system, which is {[3t 3,t, 1] T t R} (c) The kernel of L is the preimage of the zero vector, so the solution is just like the solution to (b) except with [0,0] T in place of [ 5,7] T The augmented column in the augmented matrix now consists of 0 s and, since row operations never change a column of all 0 s, we can immediately write down the reduced row echelon form of the system: [ Therefore, kerl = {[3t,t,0] T t R} ] ], 34 Linear operator A special name is given to a linear function L : R n R m in the case m = n, that is, when the domain and the codomain of L are the same: Linear operator A linear operator on R n is a linear function from R n to R n Let L be a linear operator on R 2 (the plane) Since L is a linear function from the plane to itself, we can think of it as simply moving vectors in the plane: an input vector gets moved to the corresponding output vector (A similar statement can be made for a linear operator on R n for any n) 341 Example Let L : R 2 R 2 be projection onto the x 1 -axis (a) Find the image of [2,3] T under L geometrically (b) Find the kernel of L geometrically

11 3 LINEAR FUNCTION 11 (c) Find a general formula for L(x) (d) Use the general formula found in part (c) to redo parts (a) and (b) analytically Solution (a) The image of [2,3] T under L is L([2,3] T ), which is [2,0] T : (b) The kernel of L is the set of all vectors x such that L(x) = 0 This set is the x 2 -axis, so kerl = {[0,t] T t R}: (c) A general formula for L(x) is L(x) = [x 1,0] T (keep the first component the same, but change the second component to 0) (d) Redoing part (a) using the formula, we have L([2,3] T ) = [2,0] T For part (b), we seek the set of all x for which L(x) = 0, that is, [x 1,0] T = [0,0] T This last equation forces x 1 = 0 but places no restriction on x 2, so kerl = {[0,x 2 ] T x 2 R} (which is the same as the set obtained in (b) since x 2 acts as a dummy variable, meaning that renaming it has no effect) 342 Example Let L : R 2 R 2 be 90 counterclockwise rotation

12 3 LINEAR FUNCTION 12 (a) Find the image of [3,1] T under L geometrically (b) Find the preimage of [2, 3] T under L geometrically (c) Find a general formula for L(x) (d) Use the general formula found in part (c) to redo parts (a) and (b) analytically Solution (a) The image of [3,1] T under L is L([3,1] T ), which is [ 1,3] T : (b) The preimage of [2, 3] T under L is the set of all those vectors that L moves to [2, 3] T There is only one such vector, namely [ 3, 2] T, so L 1 ([2, 3] T ) = {[ 3, 2] T }: (c) The general formula for L is L(x) = [ x 2,x 1 ] T (switch components and then negate the first) (Part (a) shows that this formula works for x in the first quadrantand one cancheckthat it worksin the otherthree quadrants as well) (d) Redoing part (a) using the formula, we have L([3,1] T ) = [ 1,3] T

13 3 LINEAR FUNCTION 13 For part (b), we seek the set of all x for which L(x) = [2, 3] T, that is, [ x 2,x 1 ] T = [2, 3] T This equation forces x 1 = 3 and x 2 = 2, so L 1 ([2, 3] T ) = {[ 3, 2] T } The functions given in the last two examples are linear (as can be checked by using the general formulas) The following functions from R 2 to itself are all linear: projection onto line through origin, rotation about origin, reflection across line through origin, dilation (= multiplication by scalar > 1), contraction (= multiplication by scalar between 0 and 1) However, translation by a vector t (which sends x to x+t) is not linear if t is nonzero (since, for instance, it does not send 0 to 0) 35 Matrix of a linear function We have seen that if A is an m n matrix, then we get a linear function L : R n R m by defining L(x) = Ax Here we turn things around and show that if we start with a linear function L : R n R m, then we can use it to build a matrix A so that the above equation holds The construction requires the following notation: In R 2, e 1 = [ [ 1 0, e 0] 2 = ; 1] In R 3, e 1 = 1 0, e 2 = 0 1, e 3 = 0 0 ; and so forth These are the standard unit vectors

14 3 LINEAR FUNCTION 14 Matrix of a linear function Let L : R n R m be a linear function There is a unique m n matrix A such that L(x) = Ax for all x R n Moreover, A = [ L(e 1 ) L(e 2 ) L(e n ) ] The matrix A is called the matrix of L This is a special case of a theorem that will be presented later, so we postpone the proof till then 351 Example Let L : R 2 R 3 be the linear function given by L(x) = x 1 +4x 2 3x 1 x 2 (a) Find the matrix of L (b) Use part (a) to find L([5, 2] T ) (c) Find L([5, 2] T ) directly from the formula for L and verify that it agrees with the answer to part (b) x 2 Solution (a) We have L(e 1 ) = L([1,0] T ) = (1)+4(0) 3(1) (0) = 1 3, (0) 0 and similarly, L(e 2 ) = [4, 1,1] T, so the matrix A of L is A = [ L(e 1 ) L(e 2 ) ] = (b) Using the formula L(x) = Ax, we have L([5, 2] T ) = 1 4 ] 3 1 [ 5 =

15 3 LINEAR FUNCTION 15 (c) The formula for L gives L([5, 2] T ) = (5)+4( 2) 3(5) ( 2) = 3 17, ( 2) 2 in agreement with part (b) 352 Example Let L : R 2 R 2 be reflection across the x 2 -axis (a) Find the matrix of L (b) Use part (a) to find L([1,3] T ) (c) Find L([1,3] T ) geometrically and verify that it agrees with the answer to part (b) Solution (a) The matrix A of L is A = [ L(e 1 ) L(e 2 ) ] = (b) Using the formula L(x) = Ax, we have L([1,3] T ) = = 0 1][ (c) Since reflection across the x 2 -axis negates the first component of a vector and keeps the second component the same, we have L([1,3] T ) = [ 1,3] T, in agreement with part (b) 36 Composition The reader is likely familiar with the concept of a composition of real-valued functions For instance, if f(x) = 2x+3 and g(x) = x 2, then the composition of f and g is given by (g f)(x) = g(f(x)) = (f(x)) 2 = (2x+3) 2 The composition can be described as doing f first and then g In more detail, the composition takes an input x, uses f to produce the output f(x), and then uses g with input f(x) to produce the final output g(f(x))

16 3 LINEAR FUNCTION 16 We can compose linear functions as well: If L : R n R m and M : R m R l are linear functions, then the composition of L and M is the function M L : R n R l given by (M L)(x) = M(L(x)) In order for the composition to make sense, the domain of M must be the same as the codomain of L (both equal to R m above), for otherwise the output produced by L could not be used as an input for M 361 Example Let L : R 2 R 3 be the linear function given by L(x) = x 1 +2x 2 x 1 3x 1 4x 2 and let M : R 3 R 2 be the linear function given by 5x1 +x M(x) = 2 7x 3 x 1 +x 3 Find a formula for the composition M L : R 2 R 2 Solution We have (M L)(x) = M(L(x)) = M x 1 +2x 2 x 1 3x 1 4x 2 5(x1 +2x = 2 )+( x 1 ) 7(3x 1 4x 2 ) (x 1 +2x 2 )+(3x 1 4x 2 ) 17x1 +38x = 2 2x 1 6x 2 Theorem Let L : R n R m and M : R m R l be linear functions, let A be the matrix of L, and let B be the matrix of M (a) M L is linear, (b) the matrix of M L is BA Proof (a) For any x,y R n, we have (M L)(x+y) = M(L(x+y)) = M(L(x)+L(y)) (L is linear) = M(L(x))+M(L(y)) (M is linear) = (M L)(x)+(M L)(y),

17 3 LINEAR FUNCTION 17 so M L satisfies the first property of a linear function Verification of the second property is left to the exercises (see Exercise 3 10) (b) For any x R n, we have (M L)(x) = M(L(x)) = M(Ax) = B(Ax) = (BA)x, so the matrix of M L is BA (by the uniqueness assertion in 35) 362 Example Let L : R 2 R 3 and M : R 3 R 2 be as in Example 361 (a) Find the matrix A of L and the matrix B of M and use these matrices to find the matrix C of the composition M L (b) Use the formula for M L found in Example 361 to find the matrix of M L directly and compare with the answer to part (a) Solution (a) We have A = [ L(e 1 ) L(e 2 ) ] = and B = [ M(e 1 ) M(e 2 ) M(e 3 ) ] = Therefore, according to the theorem, the matrix C of the composition is C = BA = = (b) Using the formula for M L found in Example 361, we have C = [ (M L)(e 1 ) (M L)(e 2 ) ] =, 2 6 in agreement with part (a) 363 Example Let L : R 2 R 2 be 90 counterclockwise rotation and let M : R 2 R 2 be projection onto the x 1 -axis (a) Using geometry, find the matrix A of L and the matrix B of M and then use these matrices to find the matrix C of the composition M L

18 3 LINEAR FUNCTION 18 (b) Using geometry, find the matrix of M L directly and compare with the answer to part (a) Solution (a) Using geometry to see where L and M send the vectors e 1 and e 2, we get A = [ L(e 1 ) L(e 2 ) ] 0 1 = 1 0 and B = [ M(e 1 ) M(e 2 ) ] = Therefore, according to the theorem, the matrix C of the composition is C = BA = = (b) The vectore 1 = [1,0] T is sent by Lto [0,1] T, which is sentby M to [0,0] T, so (M L)(e 1 ) = [0,0] T Similarly, (M L)(e 2 ) = [ 1,0] T Therefore, C = [ (M L)(e 1 ) (M L)(e 2 ) ] 0 1 =, 0 0 in agreement with part (a) 3 Exercises 3 1 Show directly from the definition that the function L : R 3 R 2 given by 2x1 x L(x) = 2 +4x 3 x 1 6x 3 is linear

19 3 LINEAR FUNCTION Show directly from the definition that the function L : R 2 R 1 given by L(x) = [ 5x 1 8x 2 ] is linear 3 3 Let L : R 3 R 2 be the linear function corresponding to the matrix A = (see Section 32 for what this means) (a) Find L([4,1, 3] T ) (b) Determine whether [6,2, 2] T is in L 1 ([4, 8] T ) (c) Find L 1 ([4, 8] T ) and use it to verify your answer to part (b) 3 4 In each case, determine whether the function F is linear: (a) F : R 2 R 1 given by F(x) = [ x 2 1 2] x2, (b) F : R 2 R 2 x1 given by F(x) = cosx Let L : R 2 R 3 be given by x 1 +3x 2 L(x) = 2x 1 6x 2 3x 1 +9x 2 (a) Find the image of [ 2,1] T under L (b) Is [ 2,4,6] T in iml? Explain 3 6 Find the kernel of the linear function L : R 3 R 3 given by L(x) = [0,2x 2 x 3, 6x 2 +3x 3 ] T

20 3 LINEAR FUNCTION Let L : R 2 R 2 be reflection across the 45 line x 2 = x 1 (In calculus, this line is written y = x) (a) Find the image of [2,1] T under L geometrically (b) Find the preimage of [1, 3] T under L geometrically (c) Find a general formula for L(x) (d) Use the general formula found in part (c) to redo parts (a) and (b) analytically 3 8 Let L : R 3 R 2 be the linear function given by 2x1 x L(x) = 2 +5x 3 7x 2 +4x 3 (a) Find the matrix of L (b) Use part (a) to find L([3,2, 1] T ) (c) Find L([3,2, 1] T ) directly fromthe formulaforlandverifythat it agrees with the answer to part (b) 3 9 Let L : R 2 R 2 be 90 clockwise rotation (a) Find the matrix of L (b) Use part (a) to find L([2,1] T ) (c) Find L([2,1] T ) geometrically and verify that it agrees with the answer to part (b) 3 10 Let L : R n R m and M : R m R l be linear functions Verify that (M L)(αx) = α(m L)(x) for all x R n and α R (This is the second part of the verification that M L is linear See the theorem of Section 36 and its proof)

21 3 LINEAR FUNCTION Let L : R 3 R 2 be the linear function given by x1 +4x L(x) = 2 +2x 3 3x 1 x 3 and let M : R 2 R 3 be the linear function given by 7x 1 +x 2 M(x) = x 1 +6x 2 5x 2 Find a formula for the composition M L : R 3 R Let L : R 2 R 2 be reflection across the line x 2 = x 1 and let M : R 2 R 2 be projection onto the x 2 -axis (a) Using geometry, find the matrix A of L and the matrix B of M and then use these matrices to find the matrix C of the composition M L (b) Using geometry, find the matrix of M L directly and compare with the answer to part (a)