Math 313 Midterm II KEY Spring 2011 sections 001 and 002 Instructor: Scott Glasgow

Transcription

1 Math 33 Midterm II KEY Spring 20 sections 00 and 002 Instructor: Scott Glasgow Write your name very clearly on this exam ooklet In this ooklet, write your mathematics clearly, legily, in ig fonts, and, most important, have a point, ie, make your work logically and even pedagogically acceptale (Other human eings not already understanding 33 should e ale to learn from your exam) To avoid excessive erasing, first put your ideas together on scratch paper, then commit the logically acceptale fraction of your scratchings to this exam ooklet More is not necessarily etter: say what you mean and mean what you say Honor Code: After I have learned of the contents of this exam y any means, I will not disclose to anyone any of these contents y any means until after the exam has closed Also, I will not even speak of this exam to anyone else in any fashion, not even whether it was difficult or not My signature elow indicates I accept this oligation Signature: (Exams without this signature will not e graded)

2 ) For the given set of ojects, together with the indicated notions of addition and scalar multiplication, determine whether each of the ten vector space axioms holds: real pairs x, y, where kx 2ky 2kx 4ky,, :,,, :, 5 5 xy x y x x y y kxy () 25pts 2 ) through 5) : Since V ut with only scalar multiplication differing, these axioms hold (since they reference only vector addition) 6) kx, y kx V when x, yv and k since oth clearly real numers then 7) We have kw z 2 5 ky and 2 kx 4 ky are k x w k y z k x w k y z kx, yw, zkxw, yz, 5 5 kx 2ky kw 2kz 2kx 4ky 2kw 4kz, kx 2ky 2kx 4ky kw 2kz 2kw 4kz,, k x, y,, so this axiom holds 8) We have k mx y so this axiom holds 9) We have 2 2 4, k m x k m y k m x k m y, 5 5 kx 2ky mx 2my 2kx 4ky 2mx 4my, kx 2ky 2kx 4ky mx 2my 2mx 4my,, k x, y,, mx y

3 mx 2my 2mx 4my kmx, y k, 5 5 mx 2my 2mx 4my mx 2my 2mx 4my k 2k 2k 4k , 5 5 (2) 5kmx 25kmy 25kmx 45kmy 5 5, km x km y km x km y, kmx, y, 5 5 so this axiom holds 0) We have x 2y 2x4y x2y 2x4y xy, :,, which is not x, yin every instance For example, ,, 0,02, Thus all axioms hold except 5 5 the last 2) rove that for any (real) vector space V,,, (satisfying the ten axioms) no matter how izarre the addition and the scalar multiplication are we must have 0u zfor any vector u V ( ), where z is the zero vector in the space, ie where z is the additive identity in V Be sure to list the axioms used in your proof Feel free to use the fact that w v vw z, or vw vw z, (3) ie feel free to use the fact that if a vector w acts like z even for just one v V, then it is z On the other hand, you may also do what you did in the relevant type of homework prolems (which invents the fact indicated in equation (3) for you) 0pts

4 By axiom 8) 0u0u 00 u, (4) which, y property of the numer 0, gives 0u 0u 0 u (5) But now this is the left-hand side of equation (3) aove with w 0 u(and, less important, v 0 u) So y the right-hand side of equation (3) w 0 u z 5pts 3) By use of the relevant if and only if theorem, determine whether the following is a suspace of M nn : ( M nn is the vector space of n nmatrices with ordinary matrix addition and scalar multiplication) the set of all n nmatrices A such that AB BAfor some fixed/specific/particular n n matrix B MAKE SURE AND REFERENCE AND USE THE THEOREM in determining your conclusion Since such A s certainly form a non-empty suset of M nn (choose A Bfor example), then, theorem, they form a suspace of M nn iff for all scalars c and c and for every A and Ain the set we have ca cab B ca ca Now (6) always holds and, so, the set is actually a suspace of (6) M nn : we have ca cab cab c AB cba cba B ca ca (7) y matrix algera, together with A and A s memership in the set (giving oth AB BAand AB BA)

5 4) Let, 2, 3,, 5pts S v v v e a asis for a vector spacev It turns out that S u u u is also a asis for V, where 2 3 u v, u v v, u v 2 v v (8) So suppose the coordinate vector S given y What is the coordinate vector S v of a vector v V relative to asis S is ac,, v (9) S v of v relative to asis S? By definition of coordinate vector, (9) holds with,, S v v v iff 2 3 v av v cv (0) 2 3 (8) is inverted y noting then that v u, v2 u2 v uu2, () v u v 2v u u 2 u u u 2u u ie (8) implies the inverse transformations v u, v2 uu2, v3 u2 u2 u 3 (2) Using (2) in (0) we get vavv2 cv3 auuu2 cu2u2 u3 (3) ac u 2 c u cu Thus, y definition, 2 3 v ac, 2 c, c (4) S 20pts 5) For the previous prolem, what is the transition matrix SS from asis S to asis S? What is the transition matrix SS from asis S to asis S? Don t mix these two up Maye the easiest thing to do here is use the result of the last prolem, ut re-inventing can e useful as an independent check By definition, the transition matrix SS from asis S to asis S ehaves as v SS S v, (5) S where the coordinate vectors are oviously taken as column vectors For prolem 4), (5) is

6 a ac SS 2 c (6) c c But, y relevant transformations, (6) gives, ac a a ac SS 2c SS 2c c c c c (7) a ac a SS 2c 2 c c 0 c 3 By making in turn choices of ac,, corresponding to standard asis elements (of ), on finds that we must have then that the columns of SS must e in turn the columns of the3 3 matrix on the right of (7), ie SS must e that matrix: one finds SS 2 (8) 0 To find SS, we note y definition it ehaves as v SS S v, (9) S which for prolem ) is evidently a ac a SS 2c 2, (20) c c 0 c and logic as aove dictates SS 2 (2) 0 One checks the matrices of (8) and (2) are inverses, and that SS, 2 SS, (22) SS , 0 which patently agrees with (8) (And this indicates an alternate, perhaps etter approach)

7 6) The eigenvalues of 2 2 A (23) are, 2, and 3 Compute the eigenspaces associated to each of these eigenvalues (Recall eigenspaces are suspaces, hence specified as the span of a asis) 5pts We have E Aker I Aker 2 ker 2 ker span, E2 Aker 2I Aker 2 2 ker ker span, E3 Aker 3I Aker 3 2 ker 0 2 ker span 2

8 7) Tell me whether the following statement is true or false (pick one), and tell me why it is true or false: There exists a 45matrix with 5 linearly independent columns 5 points Such a matrix would satisfy rank( A) dim ColA 5 But this is impossile since rank( A) dim RowAthe numer of rows of A which is 4 8) Tell me whether the following statement is true or false (pick one), and tell me why it is true or false: There exists a 45matrix with 4 linearly independent columns and whose null space is 2 dimensional 5 points Such a matrix would satisfy rank( A) dimcola 4 and have nullity 2 But this is impossile since, y theorem, rank( A) nullity( A) numer of columns of A pts 9) Determine whether the Wronskian tells you if the following sets of vectors 3 C are linearly independent If the Wronskian is inconclusive, say so in a), sin, cos 2 x x, ), x 3, 2x 4, c) x,3x 2,9x 5 We have a) sinx cosx cos x sin x cos x sin x W,sin x, cos x 0 cos x sin x sin x cos x sin x cos x 0 sinx cosx x x 2 2 cos sin 0 for some (all) value(s) of x, whence the set is linearly independent, ) x3 2 2x 4, 3, W x x x for some (all) value(s) of x, whence the set is linearly independent, and c)

9 x 3x2 9x5 Wx,3x2,9x for all values of x, whence use of the Wronskian is inconclusive (ut it is clear that the set is linearly dependent directly from the definition since 33x5 3x2 2 9x5 0 for every value of x ) n 0) Let uv, and let those vectors e orthogonal with respect to the Euclidean inner product rove the ythagorean theorem uv u v Note in your calculation the step at which you use the vectors orthogonality 5pts uv uv uv uu2uvvv u 2uv v 2 2 u 2 0 v (Orthogonality) = u 2 2 v (024) ) Three (and only three) rules fix the value of the determinant of a square matrix A : a) The determinant of the relevant identity is chosen to e ) The value of the determinant is negated y interchanging rows c) The determinant function is linear in each of A s rows individually As in the homework, the latter means that for an n nmatrix A, we always get r r r r r k r k r k r k det rk det a det a det (25) r k rk rk rk rn rn rn rn Here it is meant that the k th row vector r k, which is also the linear comination a of a row vector a and a row vector, could e from any row of the matrix, ie from any of the rows from to n Use these and only these rules to compute

10 a det (26) Hint: Realize that rule ) implies that the determinant of a matrix with two rows identical is zero 5 pts By rule c), in particular linearity in the first row, we have 0 det a adet det By rule c) again, ut now as linearity in second rows, then gives us a 0 det adet det 0 acdet ddet cdet ddet 0 0 From rule ), including the implied hint, we then have a 0 0 det a cdet 0 ddet cdet d det 0 0 ac0ddet det 0 c 0 d 0 0 ad det et ad det et 0 0 ad cdet Finally, from rule a) we get a 0 det ad cdet ad c ad c (27) (28) (29) (30)