Linear Algebra Review Session 1

Linear Algebra: Review Session 1 Vector Spaces and Subspaces Subspaces: A subset us of a ris. V that is a vector space on its aνπ. Ex. 1. (1) Suppose V is a vector space. Explain what it means to say that a subset U of V is a subspace. (2) Let VR2 and W = ((x,y) R2|6z+5y3). Determine whether or not W' is a subspace of V and justify your answer. (3) Give two more examples of subsets of R2: one that is a subspace and one that is not a subspace. Justify your answers. Ex. 2. Let W and W₂ be subspaces of a vector space V. (1) Prove that WinW₂ is a subspace of V. (2) Give an example to show that WUW, need not be a subspace of V. Ex. 3. Determine whether or not each of the following sets W is a subspace of the vector space V. Justify your answers. (1) VP(R) be the vector space of all polynomials and WpP(R) p(2) = 0}. (2) VP(R) be the vector space of all polynomials and W{pe P(R) p(2) = 1). (3) VR and W= {( x,y,z) ∈ R³²=0). (4) VM2x2(R) be the vector space of 2x 2 matrices with real coefficients and W- Max2(R) 134-2d=0}. Linear Combinations and Spans Span's) is the set of all linear combinations of vectors in MER XIESS Sie. Span (3) Caix, t... tanxn Ex. 4. Suppose that V is a vector space and u and vare vectors in V. Show that Span({u,v)) = Span((u+v, u). Ex. 5. Let S be a subset of a vector space V. Prove that Span(S) is a subspace of V. Linear Independence and Dependence A set S is linearly independent if whenever aitR and xie S and aix, t... tanxn = 0 then ai=0 for all i. Ex. 6. Suppose that u and v are vectors in a vector space V. Prove that (u, v) is linearly independent if and only if (u+vuv) is linearly independent. Ex. 7. Is the following set of polynomials linearly independent? Explain your answer. (2+2+x+2) Ex. 8. Let W₁ and W₂ be subspaces of a vector space V and assume that WinW₂ = {0}. Let w W and wye W₂ be such that w₁0 and 0. Prove that (wi, w₂) is linearly independent. Basis and Dimension A subset SEV of a vis. V is a basis of Vit Sis lin. Ind Span's)=V Ex. 9. dim (V) elements in a basis of V dim (203)=0. (1) Explain what it means to say that a subset S of a vector space V is a basis of V. (2) Give a basis for the subspace of R spanned by the vectors 27 0 4 2 0 1 2 1 2 1 2 2 You should explain how you know that your answer really is a basis; namely, you should relate your answer to the definition you gave in part (1) for full credit, (3) Give an example of a vector in Rd that is not in the subspace in part (2). Justify your answer. Ex. 10. Let W{(x,y,z) R³+2y+3=0}. (1) Prove that W is a subspace of R³. (2) Find a basis of W. Vector Spaces (1.1.1) Definition. A (real) vector space is a set V (whose elements are called vectors by analogy with the first example we considered) together with a) an operation called vector addition, which for each pair of vectors x. y EV produces another vector in V denoted x + y, and b) an operation called multiplication by a scalar (a real number), which for each vector x EV, and each scalar ER produces another vector in V denoted cx. Furthermore, the two operations must satisfy the following axioms: 1. For all vectors x, y, and z ∈ V. (x + y) + z = x + (y + z). properties of operations 2. For all vectors x and y E V. x + y = y + x. 3. There exists a vector 0 EV with the property that x + 0 = x for vectors x E V. all existence of additive identity 4. For each vector x E V, there exists a vector denoted - with the property that xx 0. existence of additive inverses Ex 5. For all vectors x and y EV and all scalars c∈ R, c(x + y) = cx+cy. 6. For all vectors x EV, and all scalars and d∈ R, (c + d)x= cx + dx. 7. For all vectors x EV, and all scalars cand d∈ R, (cd)x = c(dx). 8. For all vectors x E V. Ix = x. (1.1.6) Proposition. Let V be a vector space. Then a) The zero vector 0 is unique. b) For all x ∈ V, Ox = 0. ex.=(xg)ER² ○ー○・(x,y)=(0,0)=8 2 c) For each x ∈ V, the additive inverse x is unique. d) For all x ∈ V, and all c∈ R, (c)x=(cx). P2(IR) = x²+bx+c I a, b, CEIRS the set of polynomials wireal coetts. of degree at most 2. In P2(K), the zero vector is the constant polynomial, p(x) = 0 fe- allx. Linear Algebra Review Session I: Vector Spaces & Subspaces Ex. 1: A subset U of a vector space V is a subspace of V if i) U ≠ ∅ and ii) ∀u, v ∈ U and c ∈ R, u + v ∈ U and cu ∈ U. (closed under addition and scalar multiplication) (2) V = R^2 and W = {(x, y) | 6x + 5y = 3} Remark: Note that a subspace must always contain 0. This follows from (1) bc. by i), U ≠ ∅ so there must be a u ∈ U. Then 1u + u = u + (-u) = 0 ∈ U by ii). In R^2, 0 = (0, 0) and 6(0) + 5(0) = 0 ≠ 3. So, 0 ∉ W and W is not a subspace of R^2. (3) W1 = {(x, y) | 6x + 5y = 0} is a subspace of R^2. W2 = {(x, y) | 6x + 5y = 1} is not a subspace. As above, 0 ∉ W2 and so W2 is not a subspace. For W1, 0 ∈ W1, so W1 ≠ ∅. Let u = (u1, u2), v = (v1, v2) ∈ W1, and c ∈ R So 6u1 + 5u2 = 0 and 6v1 + 5v2 = 0. Now cu + v = (cu1 + v1, cu2 + v2) and 6(cu1 + v1) + 5(cu2 + v2) = c(6u1 + 5u2) + (6v1 + 5v2) = 0. So, W1 is a subspace of R^2. Ex. 2 Let W1 and W2 be subspaces of a v.s. V. Since W1 and W2 are subspaces, 0 ∈ W1 and 0 ∈ W2. Thus, 0 ∈ W1 ∩ W2. Let u, v ∈ W1 ∩ W2 and c ∈ R. Since W1 and W2 are subspaces, cu + v ∈ W1 and cu + v ∈ W2. Thus, cu + v ∈ W1 ∩ W2. Therefore, W1 ∩ W2 is a subspace of V. Let V = R^2 and W1 = Span((2, 0), (3, 0)) and W2 = Span((0, 1), (0, 3)). Then W1 U W2 = {(x, y) | (x, y) = (c, 0) or (x, y) = (0, c) for c ∈ R}. So, (0, 1), (1, 0) ∈ W1 U W2 but (0, 1) + (1, 0) = (1, 1) ∉ W1 U W2. Ex. 3 subspace not a subspace not a subspace subspace Ex. 4 Suppose that V is a v.s., and u, v ∈ V. Let x ∈ Span({u + v, u - v}). Then c1, c2 ∈ R s.t. x = c1(u + v) + c2(u - v). Scratchwork: Want x = d1(u + v) + d2(u - v) for some d1, d2 ∈ R. Setting d1(u + v) + d2(u - v) = c1u + c2v ⇒ (d1 + d2)u + (d1 - d2)v = c1u + c2v ⇒ d1 + d2 = c1 and d1 - d2 = c2 Solving for d1, d2: d1 = 1/2(c1 + c2) d2 = 1/2(c1 - c2) Ex. 7 Let S = {x^4, x^4 + x^3, x^4 + x^3 + x^2}. Suppose that c1x^4 + c2(x^4 + x^3) + c3(x^4 + x^3 + x^2) = 0. Then (c1 + c2 + c3)x^4 + (c2 + c3)x^3 + c3x^2 = 0. Hence, we must have c1 + c2 + c3 = 0, c2 + c3 = 0, c3 = 0, which implies that c3 = c2 = c1 = 0. Thus, S is linearly independent. Ex. 8 Let W1 and W2 be subspaces of a v.s. V and assume W1 ∩ W2 = {0}. Let w1 ∈ W1 and w2 ∈ W2 such that w1 ≠ 0 and w2 ≠ 0. (WTS: {w1, w2} is lin ind) Suppose c1w1 + c2w2 = 0 for c1, c2 ∈ R. Then c1w1 = -c2w2. Since w1 ∈ W1, c1w1 ∈ W1. Hence, -c2w2 ∈ W1. But w2 ∈ W2 and so -c2w2 ∈ W2. Thus, -c2w2 ∈ W1 ∩ W2. Since W1 ∩ W2 = {0}, we must have -c2w2 = 0. Now since w2 ≠ 0, -c2 must be 0. So, c2 = 0. Similarly, c1w1 ∈ W1 ∩ W2 and so c1w1 = 0 implies that c1 = 0 as well. Thus, {w1, w2} is linearly independent. Ex. 9 (1) A Subset S of a vector space V is a basis of V is S is linearly independent and Span(S) = V. (2) Let V = R^4, V1 = [1, -1, 1, -1]^T, V2 = [-2, 0, -1, 2]^T, V3 = [2, 1, 2, 2]^T, V4 = [2, -1, 2, 7]^T and S = {V1, V2, V3, V4}. We notice immediately that V2 = -2V1. So V2 ∈ Span{V1, V3, V4}. We row-reduce to see if this set is lin ind. [1 -1 1 -1; 0 1 -1 2; 2 1 2 2; -1 -1 -1 -1] = [1 0 2 0; 0 1 -1 2; 2 2 2 2; 0 0 0 0] R2+R1, R3-2R1, R4+R1 [1 0 2 0; 0 1 -1 2; 0 2 0 2; 0 0 0 0] R3-2R2 Since there are pivots in columns 1 and 2, V1 and V3 are lin ind and V4 ∈ Span{V1, V3}. Thus, a basis for Span(S) is {V1, V3}. Note: other possible answers are {V1, V2} and {V2, V3}. (3) The vector [0, 1, 2, -1]^T is not in Span(S). To see this, suppose that c1[1, -1, 1, -1]^T + c2[-2, 0, -1, 2]^T = [0, 1, 2, -1]^T. Then [c1 - 2c2, -c1, c1 - c2, -c1 + 2c2]^T = [0, 1, 2, -1]^T. Hence, c1 - 2c2 = 0, -c1 = 1, c1 - c2 = 2, and -c1 + 2c2 = -1. Solving this system of equations, we get c1 = -1 and c2 = -1/2. Substituting these values back into the equation, we get [-1, 1, -3/2, 3/2]^T ≠ [0, 1, 2, -1]^T. Therefore, the vector [0, 1, 2, -1]^T is not in Span(S). Ex. 10 Let W = {(x, y, z) ∈ R^3: x + 2y + 3z = 0}. (1) Note that 0 + 2(0) + 3(0) = 0. So (0, 0, 0) ∈ W. Let (x, y, z), (a, b, c) ∈ W and k ∈ R. Then x + 2y + 3z = 0 and a + 2b + 3c = 0. Now k(x, y, z) + (a, b, c) = (kx + a, ky + b, kz + c) and kx + a + 2(ky + b) + 3(kz + c) = k(x + 2y + 3z) + (a + 2b + 3c) = 0. So k(x, y, z) + (a, b, c) ∈ W. Thus, W is a subspace of R^3. (2) If (x, y, z) ∈ W then x + 2y + 3z = 0 => x = -2y - 3z. Hence, v = (-2y - 3z, y, z) = y(-2, 1, 0) + z(-3, 0, 1) for some y, z ∈ R. So, {(-2, 1, 0), (-3, 0, 1)} spans W. Since these vectors are not multiples of each other, they are linearly independent. Therefore, {(-2, 1, 0), (-3, 0, 1)} is a basis for W.