Preparation problems for the discussion sections on October 14th and 16th1. Let v =11. Find the length of v. Find a vector u in the direction of v that has length 1 .Find a vector w that is orthogonal to v.Solution: The length of v is√12+ 12=√2. Since u = av, we have to find a so that lengthof u is 1. So:√a2+ a2= 1Thus, a =1√2and we have:u =1√211.For a vector y orthogonal to v, we need to findy1y2such that0 =11·y1y2= y1+ y2One pair y1, y2that satisfies the equation is 1, −1. So the vector1−1is orthogonal to v.2. Let u1=1√211, u2=1√21−1, and v =23. Find real numbers c1, c2such thatv = c1u1+ c2u2.Solution: Since u1and u2are orthogonal (i.e. u1· u2= 0), we have that ifv = c1u1+ c2u2.for some real number c1, c2, thenu1· v = c1u1· u1+ c2u1· u2= c1u1· u1andu2· v = c1u2· u1+ c2u2· u2= c2u2· u2.Hencec1=u1· vu1· u1=23·1√2111√211·1√211=5√2.andc2=u2· vu2· u2=23·1√21−11√21−1·1√21−1=−1√2.(Note: you can also solve the systemx1u1+ x2u2= v,to find the same answer. But observe that the above approach becomes much simpler if youare working with n orthogonal vectors in Rn.)13. Let V = {abcd: a + b + c + d = 0} be a subspace of R4.(a) Find a basis for V .(b) Find a vector that is orthogonal to V .(c) Can you find two linearly independent vectors that are orthogonal to V ?Solution:(a) We have:V =abcd: a = −b − c − d=−b − c − dbcd: b, c, d ∈ R= Span−1100,−1010,−1001So−1100,−1010,−1001is a basis for V . (If you ignore the first entry, it is easy tosee that these vectors are linearly independent)(Alternatively: note that, by definition, V = Nul([1, 1, 1, 1]). And for any null space,we know how to find a basis.)(b) LetA =−1 −1 −11 0 00 1 00 0 1Then, V = Col(A) and the orthogonal complement of V is Nul(AT). We have:AT=−1 1 0 0−1 0 1 0−1 0 0 1R2→R2−R1,R3→R3−R1−−−−−−−−−−−−−−→−1 1 0 00 −1 1 00 0 −1 1R2→R2+R3,R1→R1+R2−−−−−−−−−−−−−−→−1 0 0 10 −1 0 10 0 −1 1Thus,Nul(AT) =abcd: a = b = c = d= Span1111In particular, the dimension of orthogonal complement of V is 1 and1111is orthogonalto V .(This was too much work! (But a good illustration that there is many paths to Rome.)Note again that, by definition, V = Nul([1, 1, 1, 1]). Hence, the orthogonal complementis Col([1, 1, 1, 1]T), and we immediately find the vector [1, 1, 1, 1]Tas orthogonal to V .)(Further note that V is actually defined, right away, as those vectors that are or-thogonal to [1, 1, 1, 1]T. Make sure that you can see that by writing out the innerproduct!)2(c) No, as we showed in part (b), the dimension of the orthogonal complement of V is 1 sowe cannot find two linearly independent vectors in the orthogonal complement of V .Note: a vector v is orthogonal to V if and only of v is in the orthogonal complementof V .4. Let A =1 2 14 8 21 2 5.(a) Find an echelon form U of A. What are the column spaces Col (A), Col(U )? Are theyequal?(b) Find a basis for Col (U ) and a basis for Col(A).(c) What are the row spaces Col(AT), and Col(UT). Are they equal?(d) Find a basis for the row space of A, Col(AT).(a) We have:1 2 14 8 21 2 5R2→R2−4R1,R3→R3−R1−−−−−−−−−−−−−−−→1 2 10 0 −20 0 4R3→R3+2R2−−−−−−−→1 2 10 0 −20 0 0Col(A)=Span141,282,125and Col(U)=Span100,200,1−20. They arenot equal since the third entry of any vector in Col(U) is equal to 0 and in particular,the first column of A is not in Col(U).This illustrates the fact that the column space is not preserved by row operations!(b) Since the first column and the third column are pivot columns, a basis for Col(A) is141,125; and a basis for Col(U) is100,1−20.(c) Col(AT) and Col(UT) are equal since each row of U is a linear combination of rows ofA and vice versa. We have:Col(AT) = Col(UT) = Span121,00−2The row space, on the other hand, is preserved by row operations!(d) Non-zero rows of U form a basis for Col(AT). Hence,121,00−2is a basis forCol(AT).5. Let B =1 1 0 11 0 0 1.(a) Find a basis for Nul(B).(b) Find two linear independent vectors that are orthogonal to N ul(B).(c) Is there a non-zero vector in R2orthogonal to Col(B)?3Solution: a) We bring B to reduced echelon form:1 1 0 11 0 0 1R2↔R1−−−−→1 0 0 11 1 0 1R2↔R2−R1−−−−−−−→1 0 0 10 1 0 0.Hence, vectors in N ul(B) are of the formx1x2x3x4= x30010+ x4−1001Hence, {0010,−1001} is a basis of Nul(B).b) The row space of B is orthogonal to N ul(B). Hence it is enough to find a basis of Row(B).BT=1 11 00 01 1R2→R2−R1,R4→R4−R1−−−−−−−−−−−−−−→1 10 −10 00 0.Hence, {1101,1001} is a basis of R ow(B) (in this case, we could have argued right awaythat the two vectors are independent because it is only two and they are not multiples of eachother). Thus {1101,1001} is linearly independent and each one is orthogonal to N ul(B).c) By part a) {11,10} are the pivot columns of B and hence form a basis of Col(B).Hence dim Col(B) = 2 and so R2= Col(B). Hence a vector v that is orthogonal to Col(B),is orthogonal to every vector in R2. In particular, v is
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