χ 1 χ 2 and ψ 1 ψ 2 is also in the plane: αχ 1 αχ 2 which has a zero first component and hence is in the plane. is also in the plane: ψ 1

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58 Chapter 5 Vector Spaces: Theory and Practice 57 s Which of the following subsets of R 3 are actually subspaces? (a The plane of vectors x =(χ,χ, T R 3 such that the first component χ = In other words, the set of all vectors where χ, R χ This is a subspace To show this, we need to show that it is closed under multiplication and addition Let χ and ψ lie in the plane and let α R Show α χ is also in the plane: α χ = αχ α which has a zero first component and hence is in the plane Show χ + ψ is also in the plane: χ + ψ = χ + ψ + which has a zero first component and is therefore also in the plane (b The plane of vectors x with χ = This is not a subspace The quickest way to show that a set is not a subspace is to show that the origin is not in the set Since the original does not have a first component equal to one, it is indeed not in the set (Note: a set can have the origin in it and still not be a subspace But if it doesn t have the origin in it, it is definitely not a subspace

57 s 59 (c The vectors x with χ = (this is a union of two subspaces: those vectors with χ = and those vectors with = This is not a subspace This time the origin is in the set So we have to show it is not a subspace in some other way The next easiest way to do this is to show that it is either not closed under multiplication or under addition Now, consider the vectors (,, T and (,, T Both are in the set, but (,, T + (,, T = (,, is not So, the set is not closed under addition and is thus not a subspace (d All (linear combinations of two given vectors (,, T and (2,, T This is a subspace Let x and y both be a linear combination of these two vectors Then there exist β x, γ x, β y, and γ y such that x = β x (,, T +γ x (2,, T and y = β y (,, T + γ y (2,, T Show that the set is closed under scalar multiplication: Let α R We need to show that αx is also in the set But αx = α(β x (,, T + γ x (2,, T = (αβ x (,, T +(αγ x (2,, T and hence a linear combination of the given vectors Show that the set is closed under addition: x+y = β x (,, T +γ x (2,, T + β y (,, T + γ y (2,, T =(β x + β y (,, T +(γ x + γ y (2,, T and hence a linear combination of the given vectors (e The plane of vectors x =(χ,χ, T that satisfy χ +3χ = 2 Describe the column space and nullspace of the matrices ( ( 3 A =, B = 2 3, and C = ( χ The column space of A equals all vectors α The column space of B is all of R 2 The column space of C is the origin ( 3 Let P R 3 be the plane with equation x +2y + z = 6 What is the equation of the plane P through the origin parallel to P? Are P and/or P subspaces of R 3? The plane x+2y+z = is parallel to P and goes through the origin ((,, T satisfies the equation P is not a subspace since it does not contain the origin P is a subspace (Hint: a plane that goes through the origin is always closed under multiplication and addition, and is thus a subspace 4 Let P R 3 be the plane with equation x + y 2z = 4 Why is this not a subspace? Find two vectors, x and y, that are in P but with the property that x + y is not

6 Chapter 5 Vector Spaces: Theory and Practice This is not a subspace because it does not contain the origin The vectors (4,, T and (, 4, T lie in the plane (they satisfy the equation but (4, 4, T does not (it does not satisfy the equation 5 Find the echolon form U, the free variables, and the special (particular solution of Ax = b for ( ( 3 β (a A =, b = When does Ax = b have a solution? (When 2 6 β β =? Give the complete solution Echelon form U: ( 3 β β 2β This is consistent only if β 2β 2 = In other words, when β =2β 2 Complete solution: Note that χ,, and χ 3 are free variables Thus, the complete solution has the form + α + β + γ Solving for the boxes yields the special solution (the first vector and the vectors in the null spaces (the other three vectors: β + α + β + γ -3 (b A = 2 3 6, b = β β β 2 β 3 Give the complete solution Echelon form: When does Ax = b have a solution? (When 2 β β β 2 β 3 3β when β = β 2 = and β 3 =3β Complete solution: ( ( β -2 + α Free variable: χ Solution: only

57 s 6 6 Write the complete soluton x = x p + x n to these systems: ( 2 2 2 4 5 u v w = ( 4 and ( 2 2 2 4 4 u v w = ( 4 : The echelon form for the first system is ( 2 2 2 and hence the complete solution has the form -3 2 + α -2 7 Which of these statements is a correct definition of the rank of a given matrix A R m n? (Indicate all correct ones (a The number of nonzero rows in the row reduced form of A True (b The number of columns minus the number of rows, n m False (c The number of columns minus the number of free columns in the row reduced form of A (Note: a free column is a column that does not contain a pivot True (d The number of s in the row reduced form of A False 8 Let A, B R m n, u R m, and v R n The operation B := A + αuv T is often called a rank- update Why? The reason is that a matrix of the form αuv T has at most rank one: Partition v =(ν,,ν n T Then αuv T = alphau ( ν ν n = alpha ( ν u ν n u = ( (αν u (αν n u Thus all columns are a multiple of the same vector u and thus (n columns must be linearly dependent Hence the rank is one A rank- update thus stand for updating a given matrix by adding a matrix of rank (at most one to it

62 Chapter 5 Vector Spaces: Theory and Practice 9 Find the complete solution of 3 2 6 9 5 3 3 5 x +3y +z = 2x +6y +9z = 5 x 3y +3z = 5 3 7 3 4 6 Thus the system is inconsistent and doesn t have a solution Find the complete solution of See Section 56 3 2 2 6 4 8 2 4 3 7 3 6 4 3( 7 x y z t = 3