Math 415 - Midterm 2Thursday, October 23, 2014Circle your section:Philipp Hieronymi 2pm 3pmArmin Straub 9am 11amName:NetID:UIN:Problem 0. [1 point] Write down the number of your discussion section (for instance, AD2or ADH) and the first name of your TA (Allen, Anton, Babak, Mahmood, Michael, Nathan,Tigran, Travis).Section: TA:To be completed by the grader:0 1 2 3 4 ShortsP/1 /10 /14 /10 /10 /25 /70Good luck!1Instructions• No notes, personal aids or calculators are permitted.• This exam consists of 8 pages. Take a moment to make sure you have all pages.• You have 75 minutes.• Answer all questions in the space provided. If you require more space to write youranswer, you may continue on the back of the page (make it clear if you do).• Explain your work! Little or no points will be given for a correct answer with noexplanation of how you got it.• In particular, you have to write down all row operations for full credit.Problem 1. Let V =abcd: a + 2b − c = 0.(a) [5 points] Find a basis for V .(b) [5 points] Find a basis for the orthogonal complement of V .Solution. (a) Observe that V = Nul1 2 −1 0 . The matrix is already in reducedechelon form, so we read off that a basis for V is given by−2100,1010,0001.(b) The orthogonal complement of V is V = Col1 2 −1 0 T. Hence, a basis is12−10.2Problem 2. Consider the matrixA =1 3 3 2 02 6 9 7 0−2 −6 6 8 2.(a) [7 points] Find a basis for Nul(A).(b) [5 points] Find a basis for Col(A).(c) [2 points] Determine the dimension of Col(AT) and the dimension of Nul(AT).Solution. We start with row reducing A:1 3 3 2 02 6 9 7 0−2 −6 6 8 2 1 3 3 2 00 0 3 3 00 0 12 12 2 1 3 3 2 00 0 1 1 00 0 0 0 2 1 3 0 −1 00 0 1 1 00 0 0 0 2(a) With x2and x4as free variables, we find the basis−31000,10−110.(b) By selecting pivot columns of the original matrix, we find the basis12−2,396,002.[Since these are 3 vectors, it follows that Col(A) = R3. Note that, in general, it is notpossible to select pivot columns of the echelon form to get a basis of Col(A). In thiscase, it is coincidence that the pivot columns of the echelon form B do give a basis forCol(A) (the reason is that dim Col(A) = dim Col(B), and so Col(B) = R3as well). Butgiving the basis100,010,001for Col(A) needs justification in order to get credit!](c) dim Col(AT) = dim Col(A) = 3.dim Nul(AT) = 3 − dim Col(AT) = 0.3Problem 3. [10 points] Let T : R2→ R4be the linear transformation withT11=1−102, T0−1=0100.Find the matrix A which represents T with respect to the following bases:11,1−1of R2, and1000,0−100,0010,000−2of R4.Solution.T11=1−102=1000+0−100+ 00010−000−2T1−1= T11+ 2T0−1=1−102+ 20100=1102=1000−0−100+ 00010−000−2Hence, the matrix A which represents T with respect to the given bases is1 11 −10 0−1 −1.4Problem 4. Consider the matrixA =−1 1 0 0 00 −1 1 0 01 0 −1 0 00 0 −1 1 00 0 0 −1 10 0 1 0 −1.(a) [4 points] Draw a directed graph with numbered edges and nodes, whose edge-nodeincidence matrix is A.(b) [3 points] Use a property of the graph to find a basis for Nul(A).(c) [3 points] Use a property of the graph to find a basis for Nul(AT).Solution. (a) Left to your imagination!(b) The graph is connected, so Nul(A) is 1-dimensional with basis11111.(c) The graph has two independent loops: edge1, edge2, edge3and edge4, edge5, edge6. Hence,a basis for Nul(AT) is111000,000111.5SHORT ANSWERS[25 points overall, 2-4 points each]Instructions: The following problems have a short answer. No reason needs to be given.If the problem is multiple choice, circle the correct answer (there is always exactly one correctanswer).Short Problem 1. [2 points] Let H be a subspace of R7with basis {b1, b2, b3, b4}. What isthe dimension of H?Solution. 4Short Problem 2. [2 points] Find a vector that is orthogonal to both110and111.Solution.1−10Short Problem 3. [2 points] Let A be a matrix, and let B be its row reduced echelon form.Which of the following is true for any such matrices?(a) Nul(A) = Nul(B) and Nul(AT) = Nul(BT)(b) Nul(A) = Nul(B) and Nul(AT) 6= Nul(BT)(c) Nul(A) 6= Nul(B) and Nul(AT) = Nul(BT)(d) Nul(A) 6= Nul(B) and Nul(AT) 6= Nul(BT)(e) None of these are true for all such matrices.Solution. We always have Nul(A) = Nul(B) because the null space is preserved by rowoperations.On the other hand, Nul(AT) and Nul(BT) are usually different. However, they can, bychance, be equal (take a zero matrix, for instance).Hence, the correct answer is (e) (but (b) received full credit as well).Short Problem 4. [2 points] Let A be a 4 × 3 matrix with dim Col(A) = 3. Which of thefollowing is correct?(a) The equation Ax = 0 has infinitely many solutions.(b) The equation Ax = 0 has exactly one solution.(c) The columns of A are linearly dependent.(d) The rows of A are linearly independent.(e) None of the above is correct for all such A.Solution. There are 3 columns and dim Col(A) = 3, so the columns are linearly independent.Hence, Ax = 0 has only the trivial solution, and the correct answer is (b).6Short Problem 5. [2 points] Let V be the following subspace of R5.V =x1x2x3x4x5: 3x1− x2= 0, 3x1− x3= 0What is the dimension of V ?Solution. The two equations are clearly independent. Hence, dim(V ) = 5 − 2 = 3.Short Problem 6. [3 points] Let P2be the vector
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