Linear independenceReview.•span{v1, v2,, vm} is the set of a ll linear combinationsc1v1+ c2v2++ cmvm.• span{v1, v2,, vm} is a vector space.Example 1. Is span(111,123,−113)equal to R3?Solution. The span is equal to R3if and only if the system with augmented matrix1 1 − 1 b11 21 b21 33 b3is consistent for all b1, b2, b3.• What went “wrong”? Well, the three vectors in the span satisfy−113= −3111+ 2123.Armin [email protected]• Hence, span(111,123,−113)= span(111,123).• We are going to say that the three vectors are linearly dependent because theysatisfy−3111+ 2123−−113= 0.Definition 2. Vectors v1,, vpare said to be linearly independent if the equationx1v1+ x2v2++ xpvp= 0has only the trivial solution (namely, x1= x2== xp= 0).Likewise,v1,, vpare said to be linear ly dependent if there exist coefficients x1,, xp, not all zero,such thatx1v1+ x2v2++ xpvp= 0.Example 3.•Are the vectors111,123,−113independ e nt?• If possible, fi nd a linear dependence relation among th e m.Solution.Armin [email protected] independence of matrix columns• Note that a linear depende nce relation, such as3111− 2123+−113= 0,can be written in matr ix form as1 1 −11 2 11 3 33−21= 0.• Hence, each linear dependence relation among the co lumns of a ma tr ix A corre-sponds to a nontrivia l solution to Ax = 0.Theorem 4. Let A be an m × n matr ix.The columns ofA are linearly independent.Ax = 0 ha s onl y the solution x = 0.Nul(A) = {0}A has n pivots. ( one in each column)Special ca ses• A set of a single nonzero vec tor {v1} is always linearly independent.Why?• A set of two vectors {v1, v2} is linearly inde pendent if and only if nei t h e r of th evectors is a mul tiple of the othe r .Why?• A set of ve c tors {v1,, vp} containing the zero vec tor is linearly depende n t .Why?• If a set contains more vec tors than there are entries in each vector, then the set islinearly depen dent. In other words:Any set {v1,, vp} of vectors in Rnis linearly dependent if p > n.Why?Armin [email protected] 5. With the least amount of work possible, d e cide which of the following setsof vectors are linearly independent.(a)(321,964)(b)(321)(c) columns of1 2 3 45 6 7 89 8 7 6(d)(321,964,000)Solution.A ba sis of a vector spaceDefinition 6. A se t of vectors {v1,, vp} in V is a basis of V if• V = span{v1,, vp}, and• the vectors v1,, vpare linearly independent.In other words,{v1,, vp} in V is a basis of V if and only if every vector w in V can be uniquelyexpressed asw = c1v1++ cpvp.Example 7. Let e1=100, e2=010, e3=001.Show that {e1, e2, e3} is a basis of R3. It is called the standard basis.Solution.Definition 8. V is said to have dimension p if it has a ba sis consisting of p vectors.Armin [email protected] definition makes sense because if V has a basis of p vectors, then every basis of V has p vectors.Why? (Think ofV = R3.)Example 9. R3has dimension 3. Likewise, Rnhas dimension n.Example 10. Not all vector spaces have a finite ba sis. For instance, the vector spaceof all poly nomials has infi nite dimension.Its standard basis isRecall t h at vectors inV form a basis of V if they span V and if th e y are linearlyindependent. If we know the dimension of V , we only need to check one of these twoconditions:Theorem 11. Suppose that V has dimension d.• A set of d vectors in V are a ba sis if they spa n V .• A set of d vectors in V are a ba sis if they are linearly i ndependent.Example 12. Are the fo llowing sets a basis for R3?(a)(120,011)(b)(120,011,103,−120)(c)(120,011,103)Solution.Armin [email protected] 13. Let P2be the sp ace of polynomials of degree at most 2.• What is the dimension of P2?• Is {t, 1 − t, 1 + t − t2} a basis of P2?Solution. The stand ard basis for P2isShrinking and expanding sets of vectorsWe can find a basis for V = span{v1,, vp} by discar ding, if necessary, some of thevectors in the spanning s e t.Example 14. Produce a basis of R2from the vec t orsv1=12, v2=−2−4, v3=11.Solution.Armin [email protected] 15. Find a basis and the dimension ofW =a + b + 2c2a + 2b + 4c + db + c + d3a + 3c + d: a, b, c, d real.Solution.Every set of linearly indepe ndent vectors can be ex te nded to a basis.In other words, let{v1,, vp} be linearly independent vectors in V . If V has dimension d, then wecan find vectorsvp+1,, vdsuch that {v1,, vd} is a basis of V .Example 16. ConsiderH = span100,111.• What is the dimension of this subspace of R3?• Give a basis for H, and then exten d it to a basis of R3.Solution.Armin [email protected] ng our understandingExample 17. Subspaces of R3can have dimen sion 0, 1, 2, 3.• The only 0-dimensional subspace is• A 1-dimension al subspace• A 2-dimension al subspace• The only 3-dimensional subspace isTrue or false?• Suppos e th a t V has dimension n. Then any set in V containing more than n vec t orsmust be linearly dependent.• The space Pnof po l ynomials of degree at most n has dimension n + 1.• The vector sp ace of functio n s f: R → R is infi nite-dimensional.• Consider V = span{v1,, vp}. If one of the vec tors, say vk, in the spanning set isa linear combination of t he remaining ones, then the remaining vectors sti ll span V .Armin
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