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# TAMU MATH 304 - Lect2-06web

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MATH 304Linear AlgebraLecture 16:Basis and dimension.BasisDefinition. Let V be a vector space. A linearlyindependent spanning set for V is called a basis.Equivalently, a subset S ⊂ V is a basis for V if anyvector v ∈ V is uniquely represented as a linearcombinationv = r1v1+ r2v2+ · · · + rkvk,where v1, . . . , vkare distinct vectors from S andr1, . . . , rk∈ R.Examples. • Standard basis for Rn:e1= (1, 0, 0, . . . , 0, 0), e2= (0, 1, 0, . . . , 0, 0),. . . ,en= (0, 0, 0, . . . , 0, 1).• Matrices1 00 0,0 10 0,0 01 0,0 00 1form a basis for M2,2(R).• Polynomials 1, x, x2, . . . , xn−1form a basis forPn= {a0+ a1x + · · · + an−1xn−1: ai∈ R}.• The infinite set {1, x, x2, . . . , xn, . . . } is a basisfor P, the spac e of all polynomials.Bases for RnTheorem Every basis for the vector space Rnconsists of n vectors.Theorem For any vectors v1, v2, . . . , vn∈ Rnthefollowing conditions are e quivalent:(i) {v1, v2, . . . , vn} is a basis for Rn;(ii) {v1, v2, . . . , vn} is a spanning set for Rn;(iii) {v1, v2, . . . , vn} is a linearly independent set.DimensionTheorem Any vector space V has a basis. Allbases for V are of the same cardinality.Definition. The dimension of a vector space V ,denoted dim V , is the cardinality of it s bases.Remark. By definition, two sets are of the same cardinality ifthere exists a one-to-one correspondence between their elements.For a finite set, the cardinality is the number of its elements.For an infinite set, the cardinality is a more sophisticatednotion. For example, Z and R are infinite sets of differentcardinalities while Z and Q are infinite sets of the samecardinality.Examples. • dim Rn= n• M2,2(R): the space of 2×2 matricesdim M2,2(R) = 4• Mm,n(R): the space of m×n matricesdim Mm,n(R) = mn• Pn: polynomials of degree less than ndim Pn= n• P: the space of all polynomialsdim P = ∞• {0}: the trivial vector spacedim {0} = 0Problem. Find the dimension of the planex + 2z = 0 in R3.The general solution of the equation x + 2z = 0 isx = −2sy = tz = s(t, s ∈ R)That is, (x, y, z) = (−2s, t, s) = t(0, 1, 0) + s(−2, 0, 1).Hence the plane is the span of vectors v1= (0, 1, 0)and v2= (−2, 0, 1). These vectors are line arlyindependent as they are not parallel.Thus {v1, v2} is a basis so that the dimension ofthe plane is 2.How to find a basis?Theorem Let S be a subset of a vector space V .Then the following conditions are eq uivalent:(i) S is a linearly independent spanning set for V ,i.e., a basis;(ii) S is a minimal spanning set for V ;(iii) S is a maximal linearly independent subset of V .“Minimal spanning set” means “remove any element from thisset, a nd it is no longer a spanning set”.“Maximal linearly independent subset” means “add anyelement of V to this set, and it will become linearlydependent”.Theorem Let V be a vector space. Then(i) any spanning set for V can be reduced to aminimal spanning set;(ii) any linearly independent subset of V can beextended to a maximal linearly ind epende nt set.Equivalently, any spanning set contains a basis,while any linearly independent set is cont ained in abasis.Corollary A vector space is finite-dimensional ifand only if it is spanned by a finite set.How to find a basis?Approach 1. Get a spanning set for the vectorspace, then reduce this set to a basis.Proposition Let v0, v1, . . . , vkbe a spanning setfor a vector space V . If v0is a linear combinationof vectors v1, . . . , vkthen v1, . . . , vkis also aspanning set for V .Indeed, if v0= r1v1+ · · · + rkvk, thent0v0+ t1v1+ · · · + tkvk== (t0r1+ t1)v1+ · · · + (t0rk+ tk)vk.How to find a basis?Approach 2. Build a maximal linearly independentset adding one vector at a t ime.If the vector space V is trivial, it has t he empty basis.If V 6= {0}, pick any vector v16= 0.If v1spans V , it is a basis. Otherwise pick anyvector v2∈ V that is not in the span of v1.If v1and v2span V , they constitute a basis.Otherwise pick any vector v3∈ V that is not in thespan of v1and v2.And so on. . .Problem. Find a basis for the ve ctor space Vspanned by ve ctors w1= (1, 1, 0), w2= (0, 1, 1),w3= (2, 3, 1), and w4= (1, 1, 1).To pare this spanning set, we need t o find a relationof the form r1w1+r2w2+r3w3+r4w4= 0, whereri∈ R are not all equal to zero. Equivalently,1 0 2 11 1 3 10 1 1 1r1r2r3r4=000.To solve this system of linear equations forr1, r2, r3, r4, we apply row reduction.1 0 2 11 1 3 10 1 1 1→1 0 2 10 1 1 00 1 1 1→1 0 2 101 1 00 0 01→1 0 2 001 1 00 0 01(reduced row echelon form)r1+ 2r3= 0r2+ r3= 0r4= 0⇐⇒r1= −2r3r2= −r3r4= 0General solution: (r1, r2, r3, r4)=(−2t, −t, t, 0), t ∈ R.Particular solution: (r1, r2, r3, r4) = (2, 1, −1, 0).Problem. Find a basis for the ve ctor space Vspanned by ve ctors w1= (1, 1, 0), w2= (0, 1, 1),w3= (2, 3, 1), and w4= (1, 1, 1).We have obtained t hat 2w1+ w2− w3= 0.Hence any of vectors w1, w2, w3can be dropped.For instan ce, V = Span(w1, w2, w4).Let us check whether vectors w1, w2, w4arelinearly independent:1 0 11 1 10 1 1=1 0 11 1 00 1 0=1 10 1= 1 6= 0.They are!!! It follows that V = R3and{w1, w2, w4} is a basis for V .Vectors v1= (0, 1, 0) and v2= (−2, 0, 1) arelinearly independent.Problem. Extend the set {v1, v2} to a basis for R3.Our task is to find a vector v3that is not a linearcombination of v1and v2.Then {v1, v2, v3} will be a basis for R3.Hint 1. v1and v2span the plane x + 2z = 0.The vector v3= (1, 1, 1) does not lie in the planex + 2z = 0, hence it is not a linear combination ofv1and v2. Thus {v1, v2, v3} is a basis for R3.Vectors v1= (0, 1, 0) and v2= (−2, 0, 1) arelinearly independent.Problem. Extend the set {v1, v2} to a basis for R3.Our task is to find a vector v3that is not a linearcombination of v1and v2.Hint 2. At least one of vectors e1= (1, 0, 0),e2= (0, 1, 0), and e3= (0, 0, 1) is a desired one.Let us check that {v1, v2, e1} and {v1, v2, e3} aretwo bases for R3:0 −2 11 0 00 1 0= 1 6= 0,0 −2 01 0 00 1 1= 2 6=

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