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 5 6 ShortsP/1 /? /? /? /? /? /? /? /?Good luck!1Instructions• No notes, personal aids or calculators are permitted.• This exam consists of ? 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. Consider the matrixA =1 2 2 11 1 1 10 −1 −1 0.(a) Find a basis for Nul(A).(b) Find a basis for Col(AT).(c) Determine the dimension of Col(A) and the dimension of Nul(AT).Solution:(a) We have to transform A into the row reduced echelon form:A =1 2 2 11 1 1 10 −1 −1 0R2→R2−R1−−−−−−−→1 2 2 10 −1 −1 00 −1 −1 0R3→R3−R2,R1→R1+2R2,R2→−R2−−−−−−−−−−−−−−−−−−−−−→1 0 0 10 1 1 00 0 0 0We have two free variables, x3and x4. From the row reduced echelon form we get:x1= −x4, x2= −x3Thus,Nul(A) =x1x2x3x4: x1= −x4, x2= −x3=−x4−x3x3x4: x3, x4∈ R= Span0−110,−1001(b) (Transpose of) nonzero rows of the reduced echelon form will form a basis for Col(AT),i.e., a basis for Col(AT) is:{1001,0110}Note: nonzero rows of any echelon form of A (not necessarily the row reduced echelonform) form a basis for Col(AT).(c) The dimension of Col(A) is equal to the number of pivot columns in an echelon form(in particular the row reduced echelon form) of A, So dim(Col(A)) = 2.The dimension of Nul(AT) is equal to the number of zero rows in an echelon form (inparticular the row reduced echelon form) of A, So dim(Nul(AT)) = 1.2Problem 2. Let T : R2→ R3be the linear transformation withT20=222, T0−1=010.Find the matrix A which represents T with respect to the following bases:11,1−1for R2, and200,010,00−1for R3.Solution: We have:T (11) = T (1220−0−1) =12222−010=101=12200+ 0010+ (−1)00−1T (1−1) = T (1220+0−1) =12222+010=121=12200+ 2010+ (−1)00−1Therefore,A =12120 2−1 −13Problem 3. Consider the matrixA =−1 1 0 0 0−1 0 1 0 00 1 0 −1 00 0 −1 1 00 0 0 −1 10 1 0 0 −1.(a) Draw a directed graph with numbered edges and nodes, whose edge-node incidencematrix is A.(b) Find a basis for Col(AT) by choosing a spanning tree of this graph.(This question is not relevant for the second midterm exam!)(c) Use a property of the graph (briefly explain!) to find a basis for Nul(A).(d) Use a property of the graph (briefly explain!) to find a basis for Nul(AT).Solution:(a) Our graph is:(Note that the position of vertices does not matter at all; it only matters in which waythey are connected.)(b) Rows r1, r2, r4, and r5correspond to a spanning tree of the graph. So a basis for Col(AT)is:rT1, rT2, rT4, rT5=−11000,−10100,00−110,000−11(c) Since the graph is connected, a basis for Nul(A) is:111114(d) The graph has two independent loops: edge1, −edge3, −edge4, −edge2as well as edge3, −edge6, −edge5.A basis for Nul(AT) is given by the corresponding vectors:1−1−1−100,0010−1−15Problem 4. Let V =abcd: a − b = c.(a) Write V as a span.(b) Find a basis for the orthogonal complement of V .Solution:(a) We have:V =abcd: a − b = c=b + cbcd: b, c, d ∈ R= Span1100,1010,0001(b) Let A =1 1 0 01 0 1 00 0 0 1, then we have V = Col(AT). The orthogonal complement of V(= Col(AT)) is equal to Nul(A). We haveA =1 1 0 01 0 1 00 0 0 1R2→R2−R1−−−−−−−→1 1 0 00 −1 1 00 0 0 1R1→R1+R2,R2→−R2−−−−−−−−−−−−−→1 0 1 00 1 −1 00 0 0 1From the row reduced echelon form of A, we have:Nul(A) =x1x2x3x4: x1= −x3, x2= x3, x4= 0=−x3x3x30: x3∈ R= Span−1110Therefore,−1110is a basis for the orthogonal complement of Col(AT).(Alternative quick solution, after a moment of thought:V is defined to contain all vectors [a, b, c, d]Twhich are orthogonal to [1, −1, −1, 0]T(write out the inner product!). It follows that the orthogonal complement of V isspanned by precisely this vector, and so has basis given by [1, −1, −1, 0]T.)6Problem 5. Let P2be the vector space of all polynomials of degree up to 2, and let V be thesubspace of polynomials p(t) with the property thatZ20p(t)dt = 0.Find a basis for V .
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