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Fall 2019 Math 104 - HW 1 Solutions(If you find any errors, please let your instructor know ASAP)Problem 1. If A is orthogonal, what is det A? If A is unitary, what is det A?Solution. Since A is orthogonal, by definition, this means thatAAT= I.Taking determinants of both sides of this equation we get:1 = det(I) = det(AAT) = det(A) det(AT) = det(A)2.We used the fact thatdet(A) = det(AT)and thatdet(AB) = det(A) det(B)(properties 9 and 14on p.5 of the Laub text, respectively). So, det(A)2= 1. This means thatdet(A) = ±1.The unitary case is similar: since AAH= I, we can take determinants of both sides to get:1 = det(I) = det(AAH) = det(A) det(AH) = det(A)det(A) = | det(A)|2.Here, we used the fact thatdet(AH) = det(A). Recall what the last equality means: ifz = x + iyisa complex number, withx, y ∈ R, its norm is defined to be|z| =px2+ y2. This is the Euclideandistance in the complex plane between 0 and z. The equation zz = |z|2can be derived as:zz = (x + iy)(x + iy)= (x + iy)(x − iy)= x2− (i2)y2= x2− (−1)y2= x2+ y2= |z|2.Thus, | det(A)|2= 1, so | det(A)| = 1. This means thatdet(A) is on the unit circle.Problem 2.LetU1, . . . , Uk∈ Rn×nbe orthogonal matrices. Show that the productU = U1U2· · · Ukis an orthogonal matrix.Solution.Let’s look at the casek = 2first. A matrixAis orthogonal iffAAT= ATA = I. Wecan compute this for U as:UUT= (U1U2)(U1U2)T= (U1U2)(UT2UT1)= U1(U2UT2)UT1= U1UT1= I.(Here we use the facts that multiplication is associative, that(AB)T= BTATfor any matricesA, B, and thatU1, U2are orthogonal). So we have shown that a product of two orthogonal matricesis orthogonal.In order to see this for more generalk, we will prove it by induction. Fork = 1we know thatU1is orthogonal. Assume by induction that we knowU0:= U1· · · Uk−1is orthogonal. We need toshow that this implies U:= U1· · · Ukis orthogonal. But we have:U = U1· · · Uk−1Uk= (U1· · · Uk−1) Uk= U0Uk.Now we know thatU0andUkare both orthogonal, so by the above argument we know thatU = U0Ukis orthogonal.Problem 3.LetA ∈ Rn×n. The trace ofA, denotedTr(A), is defined as the sum of its diagonalelements, i.e., Tr(A) =Pni=1aii.(a) Show that Tr(AB) = Tr(BA), even though in general AB 6= BA.(b)LetS ∈ Rn×nbe skew-symmetric, i.e.,ST= −S. Show thatTr S = 0. Then either provethe converse or provide a counterexample.Solution. (a) Let the entries of A, B be aij, bijrespectively. Then (by definition) we have:(AB)ij=nXk=1aikbkj, (BA)ij=nXk=1bikakj.Then we can compute:Tr(AB) =nXi=1(AB)ii=nXi=1nXk=1aikbki=nXi=1nXk=1bkiaik=nXk=1nXi=1bkiaik=nXk=1(BA)kk= Tr(BA).For an example of when AB 6= BA even when A, B are both square, letA = 0 10 0!, B = 0 01 0!.ThenAB = 1 00 0!, BA = 0 00 1!.But on the other hand, Tr(AB) = Tr(BA) = 1.(b)IfSis skew-symmetric, thenST= −S, so(ST)ij= Sji= −Sijfor alli, j. Takingi = j, wegetSii= −Sii, soSii= 0. This means that ifSis skew symmetric, then all of its diagonalentries are zero. In particular, the sum of its diagonal entries are zero, so Tr(S) = 0.The converse of this statement is “ifTr(S) = 0, thenSis skew-symmetric” . To see that theconverse is not true, we need to find and example of a matrixSwhich has zero trace but isnot skew-symmetric. For example, we could take:S = 0 01 0!.ThenTr(S) = 0, but since−0 6= 1,Sis not skew-symmetric. (Note: we chose anSherewhose diagonal entries are all zero. We could have chosen anSwhose diagonal entries merelyadded up to zero. Such a matrix could never be skew symmetric unless all of its diagonalentries are individually zero).Problem 4. Consider a generic 3-by-3 upper triangular matrixS =a11a12a130 a22a230 0 a33.(a) If a11, a22and a33are nonzero, show that the only solution for Ax = 0 is x = 0.(b)If eithera11= 0ora22= 0ora33= 0, then prove that the columns are linearly dependent.(Consider all three cases separately.)(c) If a22= 0, find a nonzero element in the nullspace of A.Solution. (a) The linear system Ax = 0 is given bya11x1+ a12x2+ a13x3= 0a22x2+ a23x3= 0a33x3= 0Asa11, a22, a33are nonzero, then the third equation gives usx3= 0, plugging into the secondequation givesx2= 0, similarly, pluggingx2= x3= 0into the first equation givesx1= 0.Therefore the solution for Ax = 0 is x = 0.(b)Denote the first to third column byv1, v2, v3, respectively. Then to show they are linearlydependent is equivalent to showing there exist(a, b, c) 6= (0, 0, 0)such thatav1+bv2+cv3= 0.We consider the following three cases:(1) a11= 0: In this casev1= 0, therefore1v1+ 0v2+ 0v3= 0, which proves the columnsare not linearly independent.(2) a22= 0: In this case, supposea11= 0then it is reduced to the first case. Otherwiseassuminga116= 0then we havev2=a12a11v1, thus−a12a11v1+ v2+ 0v3= 0so the columnsare not linearly independent.(3) a33= 0: In this case we may assume without loss of generality thata11, a226= 0asotherwise it is reduced to one of the previous two cases. Notice thatv3=a23a22v2+a13−a12a23a22a11v1.Therefore the columns are linearly dependent.(c)Recall that the null space ofAis defined by the solution set ofAx = 0. Therefore this is thesame as case (2) in the privous problem. When a11= 0, then (1, 0, 0) ∈ Null(A).Otherwise when a116= 0,(−a12a11, 1, 0) ∈ Null(A).Problem 5.Consider a matrixU = [u1, · · · , un]with column vectorsui∈ Rmand another matrixV = [v1, v2, · · · , vn] with column vectors vj∈ Rp. In class, I claim thatUVT= u1vT1+ · · · + unvTn. (1)Now do either (a) or (b)(a) Prove 1 holds.(b) Programming question.Solution. (a)Let the entries ofU, Vbeuij, vijrespectively. Then we have thei, j-th entry ofUVTcould be written asnXk=1uikvkj.On the other hand,ukvTkis also a matrix of sizem × p, thei, j-th element can be written asuikvkj. Summing up over k gives us the i, j-th element of u1vT1+ · · · + unvTn, which is alsonXk=1uikvkj.Hence we have provedUVT= u1vT1+ · · · + unvTn,as desired.(b) The programming part can be found in another

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