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Two important theorems in linear algebra: the first theorem states that a spanning set of vectors can be reduced to a basis, and the second theorem states that an independent set of vectors can be extended to a basis. The document also includes examples and proofs.
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THEOREM FROM LAST TIME. If v 1 , v 2 , ..., vn span W, then some subset of v 1 , v 2 , ..., vn is a basis for W. Moreover, the basis subset can be chosen so that each deleted vector vi depends on (is a linear combination of) earlier basis vectors (i.e., basis vectors v 1 , v 2 , ..., vi-1). C (^) If the rref of the matrix [v 1 | v 2 | ... | v 5 | 0 ] of the homogeneous system x 1 v 1 + x 2 v 2 + x 3 v 3 + x 4 v 4 + x 5 v 5 = 0 is v 1 v 2 v 3 v 4 v 5
1 − 1 0 2 0 0 0 0 1 − 3 0 0 0 0 0 0 1 0
x 1 = x 2 − 2 x 4 x 3 = 3 x 4 x 5 = 0 Then the homogeneous system becomes: ( x 2 − 2 x 4 ) v 1 + x 2 v 2 + ( 3 x 4 ) v 3 + x 4 v 4 + 0 v 5 = 0 Since x 2 and x 4 are arbitrary, v 2 and v 4 are dependent. To get v 2 , set x 2 =1 and x 4 =0. This makes x 3 = x 4 = x 5 =0. Thus the system becomes v 1 + v 2 =0 and v 2 =- v 1. This process always writes a dependent vector in terms of earlier vectors. The previous theorem says that spanning sets can be reduced to a basis by possibly deleting vectors. The next theorem says that an independent set can be increased to a basis by possibly adding vectors. THEOREM. If v 1 , v 2 , ..., vk are independent vectors of a space V, then v 1 , v 2 , ..., vk can be extended to a basis by adding 0 or more vectors from the standard (or any other) basis. P ROOF. Add the basis vectors to the end of the list to get v 1 , v 2 , ..., vk, s 1 , s 2 , ..., s (^) n. This set certainly spans V since the standard basis vectors do. Applying the above procedure then deletes dependent vectors to form a basis. The deleted vectors are dependent on earlier vectors. Since the v (^) i's are independent, they are not dependent on earlier vectors and so they are not deleted. Hence the resulting basis consists of the v 1 , v 2 , ..., vk plus possibly some added vectors from the standard basis. E CWhich vectors of the standard basis should be added to the set {v 1 , v 2 } = {[1,1,1], [2,1,1]} to form a basis? Add s 1 = [1,0,0], s 2 = [0,1,0], s 3 = [0,0,1], and convert to columns. Reduce the augmented matrix for the homogeneous system xv 1 + yv 2 + us 1 + vs 2 + ws 3 = 0. v 1 v 2 s 1 s 2 s 3
È.
1 2 1 0 0 0 1 1 0 1 0 0 1 1 0 0 1 0
1 0 − 1 0 2 0 0 1 1 0 − 1 0 0 0 0 1 − 1 0
Answer: Add s 2 which gives the basis {v 1 , v 2 , s 2 }. C OROLLARY. Suppose V is a vector space. D If W is a maximal set of independent vectors (i.e., it can't be extended to a larger set of independent vectors) then W is a basis. E If W is a minimal set of vectors which span V (i.e., no proper subset of W spans V) then W is a basis.
P ROOF. D By the second theorem, W can be extended to a basis. By hypothesis, no strictly larger set is independent. Hence it must be a basis. E By the first theorem, a subset of W is a basis. By hypothesus, no strictly smaller set spans V. Hence W is a basis. E THEOREM. Suppose U and W are subsets of a vector space V and suppose U has u elements and W has w elements. D If U is independent and W spans V, then u w. E If U and W are both bases, then u = w. P ROOF OF (E FROM D. If U and W are both bases, then U is independent and W spans V, hence by D u w. Similarly w u and hence u = w. E DEFINITION. The dimension n of a vector space V is the number of elements in any basis of V. If V = {0}, the dimension is 0. C OROLLARY. Suppose W is a subset of a vector space V, W has w elements and V has dimension n. D W independent implies w n ; w > n implies W is dependent. E W spans V implies w! n ; w < n implies W doesn't span V. F W independent and w = n implies W is a basis. G W spans V and w = n implies W is a basis. P ROOF. D and E follow the the theorem above. F and G follow from the corollary above. E C{t 3 , t^2 , t, 1} is a basis for P 3 , the vector space of polynomials of degree 3. Thus P 3 has dimension 4. Explain why each the following cannot be a basis for P 3. {t^3 +1, t^2 +1, t+1} {t 3 +1, t^3 1, t^2 , t, 1} Page 102 Hw 9 Answers 4(2). (c+d)u = cu du fails. 6(2). u-u = 0 for some -u fails. 8(7). V = {r>0}, uv = uv (ordinary multiplication), ru = u r^. (1) Prove uv = vu. uv = uv = vu = vu (2) u(vw) = u(vw) =u(vw) = (uv)w = (uv)w= (uv)w (3) 0 =1. Prove u 0 = u: u 0 = u1 = u. (4) - u = 1/u. Prove u-u = 0: u-u = u(1/u) = 1 = 0 (6) (c+d)u = u(c+d)^ = (u c)(u d^ ) =(cu)(du) = cu du (7) c(du) =(du) c^ = (u d) c^ =udc^ =(cd)u. (8) 1u = u 1 =u 10(7). V = {0}. 00 = 0, r0 = 0. (1) uv = 00 = 0 = 00 = vu (3) 0 = 0. Prove u 0 = u: u 0 = 00 = 0 = u (4) - u = u. Prove u-u = 0 : u-u = 00 = 0 (5) c(uv) = c(00) = c0 = 0 = 00 = cucv (6) (c+d)u = (c+d)0 = 0 = 00 = cudu (7) c(du) = c0 = 0 = (cd)u (8)1u = 0 = u 12(2). V = {xLR: x>0} uv = uv1, cu = u. The following fail: u(vw) = (uv) w, For some 0, u 0 = u , u-u = 0 for some -u, (c+d)u=cudu 14(2). V = R, uv = 2uv, cu = cu. The following fail: uv = vu, u(vw) = (uv)w For some 0, u 0 = u, u-u = 0 for some -u, (c+d)u=cudu