Review• {v1,, vp} is a basis of V if the vectors◦ span V , and◦ are independent.• The dimension of V is the nu mber of elements in a basis.• The columns of A are linearly independenteach column of A contai ns a pivot.WarmupExample 1. Find a b asis and the dimension ofW =a + b + 2c2a + 2b + 4c + db + c + d3a + 3c + d: a, b, c, d real.Solution.First, note thatW = span1203,1210,2413,0111.Is dim W = 4? N o, because the third vector is the sum of the first two.Suppose we did not noticeA =1 1 2 02 2 4 10 1 1 13 0 3 1 1 1 2 00 0 0 10 1 1 10 −3 −3 1 1 1 2 00 1 1 10 −3 −3 10 0 0 1 1 1 2 00 1 1 10 0 0 40 0 0 1 1 1 2 00 1 1 10 0 0 40 0 0 0Not a pivo t in every column, hence the 4 vectors are dependent.Armin [email protected][Not necessary her e, but:To get a relation, solveAx = 0. Set free variable x3= 1.Thenx4= 0, x2= −x3= −1 and x1= −x2− 2x3= −1. The relation is−1203−1210+2413+ 00111= 0.Precisel y, what we “noticed” to begin with.]Hence, a basis forW is1203,1210,0111and dim W = 3.It follows from the echelon form that these vectors are independent.Every set of li n e arly inde pendent vectors can be extend e d to a basis.In other words, let {v1,, vp} be linearly in dependent vectors in V . If V has dimens ion d, then wecan find vectors vp+1,, vdsuch that {v1,, vd} is a basis of V .Example 2. ConsiderH = span100,111.• Give a basis for H. What is the dimension of H?• Extend the basis of H to a basis of R3.Solution.•The vectors are independent. By defi nition, they span H.Therefore,(100,111)is a basis for H.In particular,dim H = 2.•(100,111)is not a basis for R3. Why?Because a basis forR3needs to contain 3 vectors.Or, because, for instan c e,001is not in H.So: just add this (or a ny other) missing vector!By construction,(100,111,001)is independent.Hence, this automatically is a basis ofR3.Armin [email protected] fo r column and null spacesBases f or null spacesTo find a basis for Nul(A):• find the parametric form of the solutions to Ax = 0,• express solutions x as a linear combination of vectors with the free variables ascoefficients;• these vectors form a basis of Nul(A).Example 3. Find a b asis for Nul(A) withA =3 6 6 3 96 12 15 0 3.Solution.3 6 6 3 96 12 15 0 3 3 6 6 3 90 0 3 −6 −15 1 2 2 1 30 0 1 −2 −5 1 2 0 5 130 0 1 −2 −5The s o lutions to Ax = 0 are:x =−2x2− 5x4− 13x5x22x4+ 5x5x4x5= x2−21000+ x4−50210+ x5−130501Hence, Nul(A) = span−21000,−50210,−130501.These vectors are clearly independen t.If you don’t see it, do compu te an echelon form! (p erm ute first and th ird row to the b ottom )Be tter yet: note that the first vector corresponds t o the solution withx2= 1 and the ot her freevariablesx4= 0, x5= 0. The second vector c orresponds to the solution with x4= 1 and the otherfree variabl esx2= 0, x5= 0. The third vectorHence,−21000,−50210,−130501is a basis for Nul(A).Armin [email protected] f or column s pacesRecall that the columns of A are inde pendentAx = 0 has only the trivial solution (namely, x = 0),A has no fre e variables.A basis for Col(A) is given by the pi vot columns of A.Example 4. Find a b asis for Col(A) withA =1 2 0 42 4 −1 33 6 2 224 8 0 16.Solution.1 2 0 42 4 −1 33 6 2 224 8 0 16 1 2 0 40 0 −1 −50 0 2 100 0 0 0 1 2 0 40 0 −1 −50 0 0 00 0 0 0The piv o t co lumns are the first and third.Hence, a basis forCol(A) is1234,0−120.Warning: For the basis ofCol(A), you have to take the columns of A, n ot th e columnsof an echelon form.Row operations do not preserve the column space.[For instance,10>R1↔ R201have different column spaces (of the same dimension).]Armin
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