Review for Midterm 2• As of yet unconfirmed:◦ final exam on Friday, December 12, 7–10pm◦ conflict exam on Monday, December 15, 7–10pmDir ected graphs• Go from directed graph to edge-node incidence matrix A a nd vice versa.• Basis for Nul(A) from conne cted subgraphs.For each c onnected subgraph, get a ba sis vectorx that assigns 1 to all nodes in th at subgraph,an d0 to all ot her nodes.• Basis for Nul(AT) from (independent) loops.For each (independent) loop, get a basis vect ory that assigns 1 and −1 (depending on direct ion)to the edges in that loop , and0 to all ot her edges.Example 1.Basis for N ul(A):1111Basis for N ul(AT):1−1100,00−11−1123412 3 45Fundamental no tions• Vectors v1,, vnare independent if the only linear relationc1v1++ cnvn= 0is the one with c1= c2== cn= 0.How to check for i ndependence?The columns of a matrix A are independentNul(A) = {0}.• Vectors v1,, vnin V are a basis for V if◦ they span V , tha t is V = span{v1,, vn}, and◦ they are independent.In tha t case,V has dimension n.• Vectors v , w in Rmare orthogonal if v · w = v1w1++ vmwm= 0.Ar min Strau [email protected]• From an echelon form of A, we get bases for:− Nul(A) — by solving Ax = 0− Col(A) — by taking the pivot column s of A− Col(AT) — by taking the nonzero rows of the echelon formExample 2.A =1 2 0 42 4 −1 33 6 2 224 8 0 16>RREF1 2 0 40 0 1 50 0 0 00 0 0 0Basis for Col(A):1234,0−120Basis for Col(AT):1204,0015Basis for Nul(A):−2100,−40−51Dimension of Nul(AT): 2• The solutions to Ax = b are given by xp+ Nul(A).• The fundamental theorem sta t es t h at◦ Nul(A) an d Col(AT) are orthogonal complementsSo :dim Nul(A) + dim Col(AT) = n (number of columns of A)◦ Nul(AT) and Col(A) are orthogonal complementsSo :dim Nul(AT) + dim Col(A) = m (number of rows of A)◦ In particular, if r = rank(A) (nr of pivots):− dim Col(A) = r− dim Col(AT) = r− dim Nul(A) = n − r− dim Nul(AT) = m − rExample 3. Consider the following subspac es of R4:(a) V =abcd: a + 2b = 0, a + b + d = 0Ar min Strau [email protected](b) V =a + b − cb2a + 3cc: a, b, c in RIn each case, give a basis for V and it s orthogonal complement.Try to immediately get an idea what t he d imensions are going to be!Solution.•First step: express these subspaces as one of the four subspaces of a ma tr ix.(a) V = Nul1 2 0 01 1 0 1(b) V = Col1 1 −10 1 02 0 30 0 1• Give a basis for each.(a) row reductions:1 2 0 01 1 0 1 1 2 0 00 −1 0 1 1 0 0 20 1 0 −1basis for V :0010,−2101(b) row reductions:1 1 −10 1 02 0 30 0 1 1 1 −10 1 00 −2 50 0 1(no need to continue; w e already see that the columns are independent)basis forV :1020,1100,−1031• Use t he fundam e ntal theorem to find bases for the orthogonal complements.(a) V⊥= Col1 2 0 01 1 0 1Tnote the two rows are clearly independent.basis forV⊥:1200,1101(b) V⊥= Nul1 1 −10 1 02 0 30 0 1T= Nul 1 0 2 01 1 0 0−1 0 3 1!row reductions:1 0 2 01 1 0 0−1 0 3 1 1 0 2 00 1 −2 00 0 5 1 1 0 0 −2/50 1 0 2/50 0 1 1/5basis for V⊥:2/5−2/5−1/51Ar min Strau [email protected] 4. What does it mean for Ax = b i f Nul(A) = {0}?Solution. It mea n s t hat if th e re is a sol ution, then it i s unique.That’s because all solutions toAx = b are given by xp+ Nul(A).Linear transf ormationsExample 5. Let T : R2→ R3be the linear map represented by the matrix1 02 13 0with respect to the base s01,1−1of R2and110,101,011of R3.(a) Wh at is T11?(b) Wh ich matrix represents T with respect to the standard bases?Solution.The matrix tells us that:T01= 1110+ 2101+ 3011=345T1−1= 0110+ 1101+ 0011=101(a) Note that11= 2 ·01+1−1.Hence,T11= 2T01+ T1−1= 2345+101=7811.(b) Note that10=01+1−1.Hence,T10= T01+ T1−1=345+101=446.We already know that T01=345.So, T is represented by4 34 46 5with respect to the stan dard bases.Ar min Strau [email protected] your understandingThink a bout why e ach of the se statement s i s true!• Ax = b has a solution x if and only if b is in Col(A).That’s becauseAx are linear combinations of the columns of A.• A and AThave the same rank.Recal l that the rank ofA (number of pivots of A) equals dim Col(A).So this is another way of saying thatdim Col(A) = dim Col(AT).• The columns of an n × n matrix are independent if and only if the rows are.Let r be the rank of A, and let A be m × n for now.The columns are independ entr = n (so that dim Nul(A) = 0).But also: the rows are independentr = m.In the casem = n, these two conditio ns are equivalent .• Ax = b has a solution x if and only if b is orthogonal to Nul(AT).This follows from “Ax = b has a solution x if and only if b is in Col(A)”together with the fundamental theorem, which says tha tCol(A) is the orthogonal complementof Nul(AT).• The rows of A are indep endent if and only if Nul(AT) = {0}.Recal l that element s ofNul(A) corre spond to linear relations between the columns of A.Likewise, elemen ts ofNul(AT) correspond to linear relations between the rows of A.Ar min Strau
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