bergman_notes - Math 110 Midterm Exam Professor K A Ribet...

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Math 110 Professor K. A. Ribet Midterm Exam September 26, 2002 Please put away all books, calculators, digital toys, cell phones, pagers, PDAs, and other electronic devices. You may refer to a single 2-sided sheet of notes. Please write your name on each sheet of paper that you turn in. Don’t trust staples to keep your papers together. The symbol “ R ” denotes the field of real numbers. In this exam, “0” was used to denote the vector space { 0 } consisting of the single element 0. These solutions were written by Ken Ribet. Sorry if they’re a little terse. If you have a question about the grading of your problem, see Ken Ribet for problems 1–2 and Tom Coates for 3–4. 1. (9 points) Let T : R 4 R 3 be the linear transformation whose matrix with respect to the standard bases is A = - 1 - 1 7 5 1 0 - 5 - 3 0 - 1 2 2 . (In the book’s notation, T = L A .) Find bases for (i) the null space and (ii) the range of T . The null space consists of quadruples ( x, y, z, w ) satisfying three equations of which the first two are - x - y - 7 z +5 w = 0 and x - 5 z - 3 w = 0 . It turns out that the third equation is the sum of the first two, so we can forget it. (If you don’t notice this circumstance, you’ll still get the right answer.) Replace the first equation by the sum of the first two, leaving the second alone. We get the two equations - y + 2 z + 2 w = 0 x - 5 z - 3 w = 0 . The interpretation is that z and w can be chosen freely, and then x and y are determined by z and w . If we take z = 1 , w = 0 , we get the solution (5 , 2 , 1 , 0) . With the reverse choice, we get the solution (3 , 2 , 0 , 1) . These form a basis for the null space. Once we know that the null space has dimension 2, we deduce that the range has dimen- sion 2 as well. In fact, it consists of the space of triples ( a, b, c ) with c = a + b . A basis would be the set containing (1 , 0 , 1) and (0 , 1 , 1) . Of course, there are other correct an- swers: bases aren’t unique! Let me stress that R ( T ) lies in 3-space. If your answer consists of vectors in R 4 , you’ve messed up. Note from Ribet: An answer that just has a bunch of numbers with no explanation as to what is going on is very unlikely to receive full credit. You need to tell the reader (me, in this case) what you are doing.
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2. (9 points) Let V be a vector space over a field F . Suppose that v 1 , . . . , v n are ele- ments of V and that w 1 , . . . , w n , w n +1 lie in the span of { v 1 , . . . , v n } . Show that the set { w 1 , · · · , w n +1 } is linearly dependent. Let W be the span of { v 1 , . . . , v n } . Then W is generated by n elements, so its dimension d is at most n (for example, by Theorem 1.9 on page 42). If the vectors w i were linearly independent, the set { w 1 , · · · , w n +1 } could be extended to a basis of W . This is impossible because all bases of W have d elements. 3. (10 points) Let W 1 and W 2 be subspaces of a finite-dimensional F -vector space V . Recall that W 1 × W 2 denotes the set of pairs ( w 1 , w 2 ) with w 1 W 1 , w 2 W 2 . This product comes equipped with a natural addition and scalar multiplication: ( w 1 , w 2 ) + ( w 1 , w 2 ) := ( w 1 + w 1 , w 2 + w 2 ) , a ( w 1 , w 2 ) := ( aw 1 , aw 2 ) .
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