Math 115A: Homework 4

Transcription

1 Math A: Homework page question but replace subset by tuple where appropriate and generates with spans page question but replace sets by tuple This won t be graded so do as much as you need Find bases for the following subspaces of R and tell me their dimension: x { x } W = x x R : x x x = x x { x } W = x x R : x = x = x and x + x = x (a) Does the set {x x + x x + x } span P? This should be quick! (b) Is ( ) 6 8 linearly independent in R? This should be quick! (Use question 7 from HW) Let V be a vector space over a field R and let u v w V (a) Prove that (u v) is a basis for V if and only if (u + v u v) is a basis for V (b) Prove that (u v w) is a basis for V if and only if (u + v v + w w + u) is a basis for V For each of the following prove that T is a linear transformation; calculate ker T and im T ; give a basis for ker T and im T ; say what null(t ) and rank(t ) are; check the rank-nullity theorem works out; say whether T is injective or surjective (a) T : R R defined by T (x x x ) = (x x x ) (b) T : R R defined by T (x x ) = (x + x x x ) (c) T : M M defined by ( ( ) ) x x T x = x x x ( ) x x x + x

2 6 Prove that each of the following functions T : R R is not linear (a) T (x x ) = ( x ) (b) T (x x ) = (x x ) (c) T (x x ) = (sin x ) (d) T (x x ) = ( x x ) 7 (a) Suppose T : R R is linear T ( ) = ( ) and T ( ) = ( ) What is T ( )? (b) Prove that there exists a linear transformation T : R R such that T ( ) = ( ) and T ( ) = ( ) What is T (8 )? 8 Suppose V and W are vector spaces over a field F that T : V W is a linear transformation and that v v v n V (a) Prove that if (v v v n ) is linearly dependent then (T (v ) T (v ) T (v n )) is linearly dependent (b) Prove that if (T (v ) T (v ) T (v n )) is linearly independent then (v v v n ) is linearly independent (c) Suppose ker T = {} Prove that if (T (v ) T (v ) T (v n )) is linearly dependent then (v v v n ) is linearly dependent (d) Suppose ker T = {} Prove that if (v v v n ) is linearly independent then (T (v ) T (v ) T (v n )) is linearly independent (e) Suppose im T = W Prove that if {v v v n } spans V then {T (v ) T (v ) T (v n )} spans W (f) Suppose T is an isomorphism Prove that if (v v v n ) is a basis for V then (T (v ) T (v ) T (v n )) is a basis for W (g) Prove that if {v v v n } spans V and (T (v ) T (v ) T (v n )) is linearly independent then ker T = {} (h) Prove that if {T (v ) T (v ) T (v n )} spans W then im T = W (i) Prove that if {v v v n } spans V and (T (v ) T (v ) T (v n )) is a basis for W then T is an isomorphism 9 Suppose V and W are finite-dimensional vector spaces over a field F and that T : V W is a linear transformation Use the rank-nullity to theorem to prove: (a) if dim(v ) < dim(w ) T is not surjective; (b) if dim(v ) > dim(w ) T is not injective

3 Below in each part I define a linear map T : V W give bases β V β W and an element v V You should: calculate [v] βv calculate T (v) calculate [T (v)] βw calculate [T ] β W βv confirm that [T ] β W βv [v] βv = [T (v)] βw (a) T : P P is defined by p(x) p (x) β P = ( x x x x ) β P = ( x x x x x ) v = + x x + x + x 8x Remark A polynomial is not a function by definition but you can do something which looks like a lot like differentiation to polynomials by enforcing the power rule This is a formal operation which coincides with differentiation once we allow polynomials to define a function (b) T : P P is defined by p(x) p (x) β P = (x x x x ) β P = ( x x x x x ) v = + x x + x + x 8x (c) T : P P is defined by p(x) p (x) β P = (x x x x ) β P = (x x x x x ) v = + x x + x + x 8x (d) T : P P is defined by p(x) p (x) β P = ( x x x x ) β P = ( +x +x+x +x+x +x +x+x +x +x +x+x +x +x +x ) v = + x x + x + x 8x (e) T : P P is defined by p(x) p (x) β P = ( + x + x + x + x + x + x + x + x + x + x ) β P = ( x x x x x ) v = + x x + x + x 8x (f) T : P P is defined by p(x) x β P = ( x x x x ) β P = ( x x x x x ) v = 8 + x x x + x Remark A polynomial is not a function by definition but you can do something which looks like a lot like integration to polynomials by enforcing the power rule This is a formal operation which coincides with integration once we allow polynomials to define a function (g) T : P P is defined by p(x) x β P = ( x x x x ) β P = ( x x x x x ) v = 8 + x x x + x

4 (h) In this part U = V and β U = β V A = and T : R R is defined by T (x) = Ax ( ) β R = v = Below in each part I define linear maps S : V W T : U V and give bases β U β V β W You should: calculate [T ] β V β U calculate [S] β W βv calculate ST calculate [ST ] β W βu confirm that [ST ] β W βu = [S] β W βv [T ] β V β U (a) In this part U = W and β U = β W S : P P is defined by p(x) p (x) T : P P is defined by p(x) x β P = ( x x x x ) β P = ( x x x x x ) (b) In this part U = W and β U = β W S : P P is defined by p(x) x T : P P is defined by p(x) p (x) β P = ( x x x x ) β P = ( x x x x x ) (c) In this part U = W and β U = β W S : P P is defined by p(x) p (x) T : P P is defined by p(x) x β P = ( x x x x ) β P = ( x x x x x ) (d) In this part U = W and β U = β W S : P P is defined by p(x) x T : P P is defined by p(x) p (x) β P = ( x x x x ) β P = ( x x x x x ) (a) Suppose V and W are finite-dimensional vector spaces over a field F and that T : V W is a linear transformation Show that there is a subspace U V with the following two properties: i V = ker T U; ii The linear transformation S : U im T defined by S(u) := T (u) is an isomorphism Hint: read the proof of rank-nullity and define U as the span of some set use 8(i) (b) Suppose V is a vector space over a field F and that T : V V is a linear transformation with the property that T = T Prove that V = ker T im T T means T T ie T composed with itself

5 Math A: Homework Solutions Remark I use theorem extensively throughout these solutions Together with the rank-nullity theorem it is probably the most important result we have proved so far However it doesn t have a good name so it is difficult to reference explicitly You may have to figure out for yourself when I m using it Ask for clarification if you re confused Nope R \ W so dim W < dim R = (theorem 6) β = ( ) is a linearly independent tuple in W so dim W (theorem ) Thus dim W = Since β is a linearly independent -tuple and dim W = β is a basis for W (thm ) β = ( ) is a linear independent tuple in W It also spans: suppose x x x x x W then x x x x x = x x x x x = x + x Thus β is a basis for W Use theorem again! (a) No: the set only has elements but P has dimension and < (b) No: it is a -tuple but R has dimension and >

6 Let V be a vector space over a field R and let u v w V (a) Suppose that (u v) is a basis for V Then dim V = Question 7 of homework tells us that (u + v u v) is linearly independent; since it s a -tuple and dim V = it is a basis for V (theorem ) Suppose that (u + v u v) is a basis for V Then dim V = We know from question 7 of homework that (u u) is linearly independent; since it s a -tuple and dim V = it is a basis for V (theorem ) (b) Suppose that (u v w) is a basis for V Then dim V = We know from question 7 of homework that (u + v v + w w + u) is linearly independent; since it s a -tuple and dim V = it is a basis for V (theorem ) Suppose that (u + v v + w w + u) is a basis for V Then dim V = We know from question 7 of homework that (u v w) is linearly independent; since it s a -tuple and dim V = it is a basis for V (theorem ) I ll allow myself rank-nullity to speed up my calculations (a) Let T : R R be defined by T (x x x ) = (x x x ) ( ) Let A = T is linear because (as long as we blur the distinction between row vectors and column vectors) we have T x = Ax for all x R (example ) T ( ) = ( ) and T ( ) = ( ) So im T span{( ) ( )} = R Thus im T = R (( ) ( )) is a basis for im T and rank(t ) = Rank-nullity tells us null(t ) = (( )) is a linearly independent -tuple in ker T and so it s a basis for ker T (theorem ) ker T = {(x x ) : x R} T is surjective but not injective ( and ) (b) Let T : R R be defined by T (x x ) = (x + x x x ) Let A = T is linear because (as long as we blur the distinction between row vectors and column vectors) we have T x = Ax for all x R (example ) Suppose (x x ) R and T (x x ) = This means (x + x x x ) = Thus x +x = and x x = The second equation gives x = x Substituting into the first equation we obtain x = so x = Using the first equation again we obtain x = Thus (x x ) = and ker T = {} () is a basis for ker T and null(t ) = Rank-nullity tells us that rank(t ) = (T ( ) T ( )) = (( ) ( )) is a linearly independent -tuple in im T and so it s a basis of im T (theorem ) im T = {(x x ) : x x R} T is injective but not surjective ( and ) 6

7 (c) T : M M is defined by ( ( ) ) x x T x = x x x It s linear but I m too lazy to prove that ( ( ) ( ) β ker = ( ) x x x + x ( ) ( ) ) is a linearly independent tuple in ker T so null(t ) (theorem ) ( ) ( ) β im (T ) ( ( ) ( ) ) T = is a linearly independent tuple in im T so rank(t ) (theorem ) The rank-nullity theorem forces null(t ) = and rank(t ) = so β ker is a basis for ker T and β im is a basis for im T (theorem ) { ( ) } ξ ξ ξ ker T = : ξ x x x R x x x { ( ) x x im T = T is neither injective or surjective ( and ) 6 (a) T ( ) = ( ) ( ) (b) T (( )) = T ( ) = ( 8) ( ) = T ( ) } : x x R (c) T ((π/ )) = T (π ) = ( ) ( ) = T (π/ ) (d) T (( ) + ( )) = T ( ) = ( ) ( ) + ( ) = T ( ) + T ( ) 7 (a) Suppose T : R R is linear T ( ) = ( ) and T ( ) = ( ) Then T ( ) = T (( ) ( )) = T ( ) T ( ) = ( ) ( ) = ( 8) (b) Let T : R R be defined by T (x x ) = (x x x x x x ) Then T ( ) = ( ) T ( ) = ( ) and T (8 ) = (6 ) 8 Suppose V and W are vector spaces over a field F that T : V W is a linear transformation and that v v v n V (a) Suppose that (v v v n ) is linearly dependent Then there exist λ λ λ n F not all zero such that Since T is linear this gives λ v + λ v + + λ n v n = λ T (v ) + λ T (v ) + + λ n T (v n ) = T (λ v + λ v + + λ n v n ) = T () = Since not all the λ i are zero this shows (T (v ) T (v ) T (v n )) is linearly dependent 7

8 (b) This is the contrapositive of (a) (c) Suppose ker T = {} Suppose in addition that (T (v ) T (v ) T (v n )) is linearly dependent Then there exist λ λ λ n F not all zero such that Since T is linear this gives λ T (v ) + λ T (v ) + + λ n T (v n ) = T (λ v + λ v + + λ n v n ) = λ T (v ) + λ T (v ) + + λ n T (v n ) = Thus λ v + λ v + + λ n v n ker T = {} giving λ v + λ v + + λ n v n = Since not all the λ i are zero this shows (v v v n ) is linearly dependent (d) This is the contrapositive of (c) (e) Suppose im T = W Suppose in addition that {v v v n } spans V By theorem 7 span({t (v ) T (v ) T (v n )}) = im T Thus span({t (v ) T (v ) T (v n )}) = W ie {T (v ) T (v ) T (v n )} spans W (f) Suppose T is an isomorphism Suppose in addition that (v v v n ) is a basis for V Since ker T = {} (6) and (v v v n ) is linearly independent part (d) says that (T (v ) T (v ) T (v n )) is linearly independent Since im T = W (6) and {v v v n } spans V part (e) says that {T (v ) T (v ) T (v n )} spans W Thus (T (v ) T (v ) T (v n )) is a basis for W (g) Suppose that {v v v n } spans V and (T (v ) T (v ) T (v n )) is linearly independent Let v ker T Since {v v v n } spans V we can find λ λ λ n F such that Since v ker T and T is linear we obtain v = λ v + λ v + + λ n v n = T (v) = T (λ v + λ v + + λ n v n ) = λ T (v ) + λ T (v ) + + λ n T (v n ) Since (T (v ) T (v ) T (v n )) is linearly independent this gives Thus v = and ker T = {} λ = λ = = λ n = 8

9 (h) Suppose {T (v ) T (v ) T (v n )} spans W Then W = span({t (v ) T (v ) T (v n )}) im T so im T = W (i) Suppose that {v v v n } spans V and (T (v ) T (v ) T (v n )) is a basis for W Since {v v v n } spans V and (T (v ) T (v ) T (v n )) is linearly independent part (g) says that ker T = {} Since {T (v ) T (v ) T (v n )} spans W part (h) says that im T = W Thus T is an isomorphism (6) 9 Suppose V and W are finite-dimensional vector spaces over a field F and that T : V W is a linear transformation (a) Suppose dim(v ) < dim(w ) Then dim(im T ) = rank(t ) = dim(v ) null(t ) dim(v ) < dim(w ) Thus we cannot have im T = W and T is not surjective (b) Suppose dim(v ) > dim(w ) Then dim(ker T ) = null(t ) = dim(v ) rank(t ) > dim(w ) rank(t ) Thus we cannot have ker T = {} and T is not injective (a) [ + x x + x + x 8x ] βp = ( 8) T ( + x x + x + x 8x ) = x + x + x x [T ( + x x + x + x 8x )] βp = ( ) [T ] β P P (b) [ + x x + x + x 8x ] βp = ( 8) T ( + x x + x + x 8x ) = x + x + x x [T ( + x x + x + x 8x )] βp = ( ) [T ] β P P (c) [ + x x + x + x 8x ] βp = ( 8 ) T ( + x x + x + x 8x ) = x + x + x x [T ( + x x + x + x 8x )] βp = ( ) [T ] β P P 9

10 (d) [ + x x + x + x 8x ] βp = ( 8) T ( + x x + x + x 8x ) = x + x + x x [T ( + x x + x + x 8x )] βp = ( ) [T ] β P P (e) T : P P is defined by p(x) p (x) β P = ( + x + x + x + x + x + x + x + x + x + x ) β P = ( x x x x x ) v = + x x + x + x 8x [ + x x + x + x 8x ] βp = ( 8) T ( + x x + x + x 8x ) = x + x + x x [T ( + x x + x + x 8x )] βp = ( ) [T ] β P P (f) T : P P is defined by p(x) x β P = ( x x x x ) β P = ( x x x x x ) v = 8 + x x x + x [8 + x x x + x ] βp = (8 ) T (8 + x x x + x ) = 8x + x x x + x [T (v)] βp = ( 8 ) [T ] β P P (g) T : P P is defined by p(x) x β P = ( x x x x ) β P = ( x x x x x ) v = 8 + x x x + x [8 + x x x + x ] βp = (8 ) T (8 + x x x + x ) = + 8x + x x x + x [T (v)] βp = ( 8 )

11 [T ] β P P (h) [( )] βr = ( ) T ( ) = ( ) [T ( )] βr = ( ) [T ] β R β R = (i) This wasn t a question but I wanted to give you more examples of the coordinate change matrix In (a) we calculated the matrix of T : P P defined by p(x) p (x) with respect to the standard basis for P and P Then in questions (b-e) we calculated the matrix with respect to different bases Here are the relevant versions of the formula A = P AQ: = = = =

12 In (h) we learn that = (a) ST = P [ST ] P P = I (b) (ST )(p(x)) = p(x) p() I = [ST ] P P = = (c) ST = P [ST ] P P = I (d) (ST )(p(x)) = p(x) p() I = [ST ] P P = =

13 (a) Suppose V and W are finite-dimensional vector spaces over a field F and that T : V W is a linear transformation Choose a basis (v v n ) for ker T and extend it to a basis (v v n v n+ v m ) of V Let U = span({v n+ v m }) and define S : U im T by S(u) := T (u) By definition {v n+ v m } spans U and we know from the proof of the rank-nullity theorem that (T (v n+ ) T (v m )) is a basis for im(t ) Thus 8(i) tells us that S is an isomorphism Moreover we have (homework (c)) ker T + U = span({v v n }) + span({v n+ v m }) = span({v v n } {v n+ v m }) = span({v v n v n+ v m }) = V Finally let u ker T U Then S(u) = T (u) = Since S is injective this gives u = Thus ker T U = {} and so ker T U = V (b) Suppose V is a vector space over a field F and that T : V V is a linear transformation with the property that T = T Suppose v ker T im T Since v im T we can find a v V such that v = T v Since v ker T T v = Thus v = T v = T v = T (T v ) = T v = and we conclude that ker T im T = {} Given v V we have v T v ker T because T (v T v) = T v T (T v) = T v T v = T v T v = We also have T v im T Thus v = (v T v) + T v ker T + im T and so V = ker T + im T We conclude that V = ker T im T