Review• Vectors v1,, vpare linearly dependent ifx1v1+ x2v2++ xpvp= 0,and not all the coefficients are zero.• The columns of A are linearly independenteach column of A contains a pivot.• Are the vectors111,123,−113independent?1 1 −11 2 11 3 3 1 1 −10 1 20 2 4 1 1 −10 1 20 0 0So: no, they are dependent! (Coeff’s x3= 1, x2= −2, x1= 3)• Any set of 11 vectors in R10is linearly dependent.A basis of a vector spaceDefinition 1 . A set of vect ors {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 2. Let e1=100, e2=010, e3=001.Show that{e1, e2, e3} is a basis of R3. It is called the standard basis.Solution.•Clearly, span{e1, e2, e3} = R3.• {e1, e2, e3} are inde pendent, because1 0 00 1 00 0 1has a pivot in each column.Definition 3. V is said to have dimension p if i t has a basis 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.)A basis ofR3cannot have more than 3 vectors, because any set of 4 or more vectors in R3is linearlydepen dent.A basis of R3cannot have le ss than 3 vectors, because 2 vectors span at most a plane (challenge :can you think of an argument that is more “rigorous”?).Example 4. R3has dimension 3.Indeed, the standard basis100,010,001has three elements.Likewise,Rnhas dimension n.Example 5. Not all vector spaces have a finite basis. For instance, the vector space ofall polynomi a ls has infinite dimension.Its standard b asis is1, t, t2, t3,This is indeed a basis, because any polynomial can be written as a uniqu e linear combination:p(t) = a0+ a1t ++ antnfor some n.Recall that vectors inV form a basis of V if they span V and if they are linearlyindependent. If we know the dim e nsion ofV , we only need to check one of these twoconditions:Theorem 6. Su p pose tha t V ha s dimension d.• A set of d vectors in V are a basis if they span V .• A set of d vectors in V are a basis if they are linearly independe nt.Why?• If the d vectors were not independent, then d − 1 of them w ould still span V . In the end, wewould find a basis of less thand vectors.• If the d vectors would not span V , then we could add another vector to the set and have d + 1independent ones.Example 7. Are the following sets a basis for R3?(a)(120,011)No, the set has less t han 3 elements.(b)(120,011,103,−120)No, the set has more than 3 elements.Armin [email protected](c)(120,011,103)The set has 3 elements. Hence, it is a basis if and only if the vectors are independent.1 0 12 1 00 1 3 1 0 10 1 −20 1 3 1 0 10 1 −20 0 5Since each column contains a pivot, the three vectors are independent.Hence, th is i s a basis ofR3.Example 8. Let P2be the space of polynomials of degree at most 2.• What is the dimension of P2?• Is {t, 1 − t, 1 + t − t2} a b a sis of P2?Solution.•The standard basis for P2is {1, t, t2}.This is indeed a basis because every polynomiala0+ a1t + a2t2can clearly be written as a linear com bination of 1, t, t2in a unique way.Hence,P2has dimension 3.• The set {t, 1 − t, 1 + t − t2} has 3 elements. Hence, it is a basis if and o n ly if thethree polynomials are linearly independent.We need to check whetherx1t + x2(1 − t) + x3(1 + t − t2)(x2+x3)+(x1− x2+x3)t− x3t2= 0has only the trivial soluti o n x1= x2= x3= 0.We get the equationsx2+ x3= 0x1− x2+ x3= 0−x3= 0which clearly only have the trivial solution. (If you don’t see it, solve the system!)Hence,{t, 1 − t, 1 + t − t2} is a basis of P2.Armin [email protected] and expanding sets of vectorsWe can find a basis for V = span{v1,, vp} by discarding, if necessary, some of thevectors in the spannin g set.Example 9. Produce a b asis of R2from the vectorsv1=12, v2=−2−4, v3=11.Solution. Three vectors in R2have to be line arly dependent.Here, we notice that v2= −2v1.The remain ing vectors{v1, v3} are a basis of R2, because th e two vec tors are clearlyindependent.Che ck ing our understan dingExample 10. Subspaces of R3can have dimension 0, 1, 2, 3.• The only 0-dimensional subspace is {0}.• A 1-dimensional subspace is of the form span{v } where v0.These sub spaces are lines through the origin.• A 2- d imensional subs pace is of the form span{ v , w } where v and w are notmultiples of each other .These sub spaces are planes through the origin.• The only 3-dimensional subspace is R3itself.True or false?• Suppose that V has dime n sion n. Then any set in V containing more than n vectorsmust be linearly dependent.That’s correct.• The space Pnof polynomials of degr e e at most n has dime n sion n + 1.True, as well. A basis is{1, t, t2,, tn}.• The vector space of functions f : R → R is infini te -dimensional.Yes. A still-infinite-dimensional subspace are the polyn omials.• Consider V = span{v1,, vp}. If one of the vectors, say vk, in the spanning set isa linear combination of the remaining ones, then the remaining v e c tors still spanV .True,vkis not adding any thing new.Armin
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