Chapter 2 vector space

Vector spaces Phan Thi Khanh Van E-mail: khanhvanphan@hcmut.edu.vn May 13, 2021 (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Table of contents Vector space Linear combination, linear dependence, linear independence Spanning set, basis, dimension, coordinates Coordinates of a vector with respect to a basis, Change of Coordinates Matrix Subspaces (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Vector space A vector space over a field K is a set V together with two operations: vector addition +: x + y , x, y ∈ V scalar multiplication × : αx, x ∈ V , α ∈ K , (R, C) and that satisfy the eight axioms listed below: x +y =y +x x + y + z = x + (y + z) ∃0 ∈ V : + x = x ∀x ∈ V , ∃(−x) ∈ V : (−x) + x = (α + β)x = αx + βx α(x + y ) = αx + αy α(βx) = (αβ)x 1.x = x An element x ∈ V is called a vector Examples Rn = {(x1 , x2 , xn ) : xi ∈ R} with ordinary +, × Mm×n (R) with matrix addition + and scalar multiplication × Pn (R) = {an x n + + a1 x + a0 , ∈ R} with ordinary + and × (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Linear combination, linear span of a vector set In a vector space V , let M = {e1 , em } be a vector set x is called a linear combination of M if ∃α1 , α2 , αm ∈ K : x = α1 e1 + α2 e2 + + αm em The linear span of M, which is denoted by span{M} or M , is the collection of all linear combinations of M: span{M} := {α1 e1 + α2 e2 + + αm em , ∀αi ∈ K} (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Example In R : M = {(1; 1), (2; 3), (1; −2)} Is the vector x = (3, 4) a linear combination of M? We have (3; 4) = 1.(1; 1) + 1.(2; 3) + 0.(1; −2) ⇒ (3; 4) is a linear combination of M, (it means that x ∈ span{M}) Example In a vector space V let x and y be two vectors and M = {x − 2y , 2x + 3y } Is the vector u = 4x − y a linear combination of M? We have that u = 4x − y = 2.(x − 2y ) + 1.(2x + 3y ) ⇒ u is a linear combination of M, (u ∈ span{M}) (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Example In R : M = {(−1; 2; −3), (2; −2; 1)} Is u = (5; −2; −6) a linear combination of M? Consider the equation α(−1; 2; −3) + β(2;  −2; 1) = −α + 2β =  (5; −2; −6) ⇔ 2α − 2β = −2   −3α + β = −6   α = ⇔ β=4 ⇒ α, β   −3α + β = −6 Then, u is not a linear combination of M (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Linear independence, linear dependence The set M is called linearly independent (LI) if α1 e1 + α2 e2 + + αm em = ⇒ αi = 0, ∀i: the equation has only trivial solution M is called linearly dependent (LD) if it is not linearly independent ∃(α1 , α2 , αm ) = (0, 0) satisfying α1 e1 + α2 e2 + + αm em = the equation has nontrivial solution Rank of a vector set The rank of a vector set S in the space V is the maximum number of linearly independent vectors which can be chosen from S Properties If α = 0, then r (x1 , x2 xm ) = r (αx1 , x2 xm ) For ∀α: r (x1 , x2 , xm ) = r (x1 + αx2 , x2 xm ) If x is a linear combination of a vector set M ⇔ r (M) = r (M, x) (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Properties If M ⊂ N, then rank(M) ≤ rank(N) rank(M ∪ N) ≥ rank(M), rank(N) If A ∈ Mm×n (K ) then, rank(A) = rank(rowA) = rank(colA) M is LI ⇔ r (M) = m (the number of vectors of the set) If r (M) < m then M is LD (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Example In R3 , given M = {(1, 2, −3), (4, 0, 2), (−2, −4, 6), (−2, 4, −8)} Find the rank of M and one maximal independent subset   −2 −2 A = [e1 e2 e3 e4 ] =  −4  −3 −8   −2 −2 r2 →r2 −2r1 , r3 →r3 +3r1  −−−−−−−−−−−−−−→ 0 −8 0 14 −14   −2 −2 r3 →4r3 +7r2 8 −− −−−−−→ 0 −8 0 0 r (A) = ⇒ r (M) = Maximal independent subset: {(1, 2, −3), (4, 0, 2)} (Phan Thi Khanh Van) Vector spaces May 13, 2021 / 37 Example In V given a LI set M = {x, y , z} Find the rank of N = {2x − y , x + y + z, −2z} We have r {2x − y , x + y + z, −2z} v 2→v 2+ v23 = = v 1→v 1+v r {2x − y , x + y , −2z} r {3x, x + y , −2z} v 2→v 2− v31 = r {3x, y , −2z} = r {x, y , z} = (because M is LI, then rank(M) = the number of vectors) (Phan Thi Khanh Van) Vector spaces May 13, 2021 10 / 37 Coordinates of a vector with respect to a basis Let V be an n dimensional vector space and let E = {e1 , e2 , en } ⊂ V be an ordered basis of V Then, ∀x ∈ V there is a unique linear combination of E that equals x: x = α1 e1 + α2 e2 + + αn en The sequence of coordinates (α1 , α2 , αn )T is called the coordinate vector of x relative to (with respect to) E Denote: [x]E Properties of coordinate vectors Let E be a basis of a vector space V and let x, y be two vectors in V x = y ⇔ [x]E = [y ]E [x + y ]E = [x]E + [y ]E [αx]E = α[x]E (Phan Thi Khanh Van) Vector spaces May 13, 2021 23 / 37 Example Find the coordinates of x = (1, 2, 4)T with respect to the basis E = {(1, 1, 2)T , (1, 1, 0)T , (1, 0, 0)T } of R3         1 1 x = 2 = α 1 + β 1 + γ 0 0       α + β + γ = α = ⇔ α+β =2 ⇔ β=0 , so [x]E =       −1 2α = γ = −1      1 1 α − nd way: [x] = 2 = 1 0 β  = E [x]E 0 γ Because E is a basis of R , E is invertible   −1   1 1 −1      1 ⇒ [x]E = E [x] = = 0 0 −1 (Phan Thi Khanh Van) Vector spaces May 13, 2021 24 / 37 Example Let E = {(1, 1, 2), (1, 1, 0), (1, 0, 0)} and F = {(1, 2, 3), (2, −4, 1), (−1, 1, 1)} be two bases of R3 and [x]E = (2, −2, 3)T Find [x]F [x]E = (2, −2, 3)T ⇒ x = 2.e1 − 2e2 + 3e3 = 2(1, 1, 2) − 2(1, 1, 0) + 3(1, 0, 0) = (3, 0, 4) T Let [x]F  = (α, β, = α[e  γ)  , we  have  [x]  1 ] + β[e2 ] + γ[e3 ] −1 [x] = 0 = α 2 + β   + γ   −1    23  −1 17  −4  ⇒ [x]F =   17 −10 1 17 (Phan Thi Khanh Van) Vector spaces May 13, 2021 25 / 37      1 − nd way: [x] = E [x]E = 1 0 −2 = 0 0 −1    23   −1 17  −1    0 =  17 [x]F = F [x] = −4 10 1 − 17 Remark: In Rn , let E and F be two bases, x be a vector [x] = E [x]E = F [x]F ⇒ [x]F = F −1 [x] = F −1 E [x]E TE →F = F −1 E = [e1 ]F [e2 ]F [en ]F - change of basis matrix (or transition matrix) from E to F And we have the relation between [x]F and [x]E as follows [x]F = TE →F [x]E = F −1 E [x]E (Phan Thi Khanh Van) Vector spaces May 13, 2021 26 / 37 Change of basis matrix (transition matrix) Let E = {e1 , e2 , en } and F = {f1 , f2 , fn } be bases of a vector space V The change of basis matrix from E to F is the matrix whose columns are coordinate vectors of fi , i = 1, n, with respect to the basis E : TEF = [e1 ]F [e2 ]F · · · [en ]F And TEF satisfies [x]F = TEF [x]E If V = Rn , then [ei ]F = F −1 [ei ] Hence, TEF = F −1 [e1 ] F −1 [e2 ]F F −1 [en ] = F −1 E Properties In a vector space V , let E , F and G be bases TE →F is the change of basis matrix from E to F then TE−1 →F is the change of basis matrix from F to E TG →F is the change of basis matrix from G to F , TF →E is the change of basis matrix from F to E , then TF →E TG →F is the change of basis matrix from G to E [x]E = TF →E [x]F = TF →E TG →F [x]G = TG →E [x]G (Phan Thi Khanh Van) Vector spaces May 13, 2021 27 / 37 Example In V let E = {x, y , z} be a basis Prove that F = {x − y , y − z, z − 2x} is also a basis of V Given that [u]E = (1, −2, 3)T Find [u]F −2 = ⇒ F is |F | = −1 −1  [u]F = F −1 E [u]E = −1 −1 a basis of V −1     −3 −2  −2 = −5 −2 Example In V let E , F , G be bases −1 [u]E = (1, −2)T , TE →F = , TG →F = Find [u]G 1 [u]G = TF →G TE →F [u]E = (Phan Thi Khanh Van) −1 −1 1 Vector spaces −2 May 13, 2021 28 / 37 Subspaces In a vector space (V , ×, +), a subset F with two operations ×, + which is also vector space is called a subspace of V We have: F ⊂ V ⇒ dim(F ) ≤ dim(V ), ”=” occurs if and only if F = V Theorem A nonempty subset F of a vector space V is a subspace of V if and only if ∀x, y ∈ F : x + y ∈ F ∀x ∈ F , α ∈ K : αx ∈ F Or we can use one equivalent condition: ∀α, β ∈ K , ∀x, y ∈ F : αx + βy ∈ F (Phan Thi Khanh Van) Vector spaces May 13, 2021 29 / 37 Example Prove: F = {A ∈ M2 [R] : A = 0} is a subspace of M2 [R] −1 −3 −1 −3 For all and A, B ∈ F , we have: A.C = 0, B.C = Hence, ∀α, β ∈ R : (αA + βB).C = αAX + βBC = ⇒ (αA + βB) ∈ F ⇒ F is a subspace of M2 [R] Put C = Example Prove: F = {A ∈ M2 [R] : A = a11 , aij ∈ R} is a subspace of a21 a22 M2 [R] a11 b , B = 11 a21 a22 b21 b22 αa11 + βb11 Hence, ∀α, β ∈ R, (αA + βB).C = αa21 + βb21 αa22 + βb22 ⇒ (αA + βB) ∈ F ⇒ F is a subspace of M2 [R] For all A, B ∈ F , we have: A = (Phan Thi Khanh Van) Vector spaces May 13, 2021 30 / 37 Linear span Let M = {v1 , v2 , , vp } be a vector set in a vector space V The linear span of M: U = span(M) = {x = α1 v1 + + αp vp , ∀αi ∈ R(C)} is also a subspace of V and dim(U) = rank(M) Example Let M = {(1; 2; −1; 3), (−2; 3; 1; 2), (0; 7; −1; 8)} be a vector set in R4 ⇒ Span(M) is a subspace of R4 Finding a basis of a subspace of Rn Let U be the span of a vector set M in Rn : U = span(M) = {v1 , v2 , , vp } (M is a spanning set of U) A basis of U is a maximal independent subset of M: A = v1 v2 vp Using the elementary row operations to reduce A to the echelon form rank(M) = rank(A) = dim(U) The vectors that are corresponding to the leading entries form a basis of U (Phan Thi Khanh Van) Vector spaces May 13, 2021 31 / 37 Example M = {(1, −2, 0, 1), (2, 3, 7, 5), (−1, 1, −1, 0), (3, 2, 8, 5)} Find the dimension and one basis of the span F of M     −1 −1   −2 2  → 0 −1 8 A = v1 v2 v3 v4 =  0 −1 8  −1 8 −1 −1 5   −1 0 −1 8  → 0 0 0 0 0 One basis of F : {(1, −2, 0, −1), (−2, 3, 1, 2)}, dim(F ) = (Phan Thi Khanh Van) Vector spaces May 13, 2021 32 / 37 Example In P2 (x) let F be the span of a vector set M: F = span(M) = span(x + 2x + 1, x − x − 2, x + x) Find the dimension, a basis of F A basis of F is the maximal independent M    subset of    1 1 1 1 A = v1 v2 v3 = 1 −1 2 → 0 −3 −1 → 0 −3 −1 −3 −1 0 −2 One basis of F : {x + 2x + 1, x − x − 2}, dim(F ) = (Phan Thi Khanh Van) Vector spaces May 13, 2021 33 / 37 Nullspaces In Rn given a linear homogeneous system Amxn X = The set of all solutions to the system is a subspace of Rn Denote: null(A) Example Let F be a subset in R3 : F = {(x1 , x2 , x3 ) ∈ R3 |x1 − 2x2 + x3 = ∧ x1 − x3 = 0} Prove that F is a subspace R3 Find the dimension, a basis of F Let x(x1 , x2 , x3 ) and y (y1 , y2 , y3 ) be solutions of the system, i.e x1 − 2x2 + x3 = y1 − 2y2 + y3 = , x1 − x3 = y1 − y3 = (αx1 + βy1 ) − 2(αx2 + βy2 ) + (αx3 + βy3 ) = (αx1 + βy1 ) − (αx3 + βy3 ) = αx + βy is also a solution So F is a subspace of R3 For all α, β ∈ R: (Phan Thi Khanh Van) Vector spaces , or May 13, 2021 34 / 37 To find one basis of F , we first solve the system: ∀x ∈ F : Ax = −2 1 −2 ∼ −1 0 −2    x1 = m −1 ∼ ⇔ x2 = m ⇔ x = m 1 −1   x3 = m Consequently, ∀x ∈ F : x = m(1, 1, 1)T Then {(1, 1, 1)T } is one basis of F , dimF = (Phan Thi Khanh Van) Vector spaces May 13, 2021 35 / 37 Example In R4 given F = {(x1 , x2 , x3 , x4 ) ∈ R4 |2x1 − x2 + 3x3 + x4 = 0; x1 − 2x2 + x3 = 0; 4x1 − 5x2 + 5x3 + x4 = 0} Find the dimension, one basis of F ∀x ∈ F :Ax [A|0] =  =   1 −1 −2 0  →  −2 −5 0 1  x1 = −5m − 2n    x = −m − n −2 0 → ⇔ 1 0  x  = 3m   x4 = 3n     −2 −5 −1 −1    Therefore, ∀x ∈ F : [x] = m    + n   Then {(−5, −1, 3, 0), (−2, −1, 0, 3)} is one basis of Remark: dim(null(A)) = n − rank(A) (Phan Thi Khanh Van) Vector spaces  0  F , dimF = May 13, 2021 36 / 37 Exercises In R4 given F = span{(1, 1, −1, 1), (2, 0, 3, 1), (0, −2, 5, −1), (3, 1, 2, 2)} Find one basis and the dimension of F In  R , given a subspace V - the set of solutions to the system 3x1 + x2 + x3 =  2x1 + 2x2 − x3 = Find one basis and the dimension of V   x1 + 3x2 − 3x3 = In R3 given M = {(1, 1, 1), (2, 3, 1), (0, 1, 2)} a) Is x = (1, −2, 3) in < M > ? b) Find m such that (1, 0, m) ∈< M > In R4 given F =< (1, 1, 1, 1), (1, 2, 3, 4), (3, 1, 2, 1) >, G =< (−1, 2, 2, 4), (−1, 1, m, 0) > Find m such that G ⊂ F In R4 , given a subspace V - the set of all solutions to the system   x1 + x2 − 2x3 = 2x1 + 3x2 − x3 − x4 = Find m such that dim(V ) is max With   x1 + x2 − mx4 = m(Phan found, find one basis of V Vector spaces Thi Khanh Van) May 13, 2021 37 / 37 ... subspace of M2 [R] Put C = Example Prove: F = {A ∈ M2 [R] : A = a11 , aij ∈ R} is a subspace of a21 a 22 M2 [R] a11 b , B = 11 a21 a 22 b21 b 22 αa11 + βb11 Hence, ∀α, β ∈ R, (αA + βB).C = αa21... βb21 αa 22 + βb 22 ⇒ (αA + βB) ∈ F ⇒ F is a subspace of M2 [R] For all A, B ∈ F , we have: A = (Phan Thi Khanh Van) Vector spaces May 13, 20 21 30 / 37 Linear span Let M = {v1 , v2 , , vp } be a vector. .. Van) Vector spaces May 13, 20 21 / 37 Example In R : M = {(−1; 2; −3), (2; ? ?2; 1)} Is u = (5; ? ?2; −6) a linear combination of M? Consider the equation α(−1; 2; −3) + β (2;  ? ?2; 1) = −α + 2? ? =