Ari --- True/False. If ABx = c has infinitely many solutions, then so too does Ax = c. Answer: FALSE. To get full credit, you needed a complete counterexample: two matrices A and B and a vector c such that ABx = c has infinitely many solutions and Ax = c does not. An example of a good, simple answer is, "Let A = identity, B = square of zeroes, and c be the zero vector. Then ABx = c has infinitely many solutions, since any x will do, but Ax = c only when x = 0." One frequent mistake was making an argument for the existence of a suitable counterexample without actually providing it. Unless your argument was sufficiently specific that it constituted a proof, this was worth anywhere from 4-9 points. If your argument was completely incorrect or irrelevant to the construction of a counterexample, you got 3. Mentioning that B should have linearly dependent columns bumped that up to 4, and adding that A should have independent columns made it 5. To get higher than 5, you needed some sort of counterexample or a foolproof method of constructing one. Another common mistake was omitting the vector c. Providing a correct A and B was usually 8 points. Note that the fact that AB has nontrivial nullspace does _not_ imply that ABx = c always has infinitely many solutions; it could have zero, depending on your choice of c. Another was letting A be any nonzero square matrix, B the zero matrix, and c the zero vector. This only works if A has trivial nullspace; otherwise, Ax = c can have infinitely many solutions even though A is nonzero. Not all nonzero matrices are invertible! It is _not_ true that Ax = 0 implies A = 0 or x = 0! Other mistakes included omitting A or B, multiplying A and B incorrectly, getting the sizes of matrices wrong, doing elementary row operations to one side of an equation and not the other, assuming that x was the same in both equations, and a variety of other shenanigans. I was not especially charitable in grading proof attempts, because the instructions specifically called for a counterexample. In particular, unless your argument was actually a step towards a solution, you got nothing, even if you said something completely true. Yuan/Alice ---------- True/False. Let V be the graph of a linear transformation M. Then V is a subspace. Answer: TRUE. Two correct answers: 1) Check that V is closed under the vector addition and scalar multiplication; 2) or, simply write V as the span of two independent vectors (1,0,m3,m4)^T and (0,1,m3,m4)^T (where denotes M by (m1,m2;m3,m4)). Many kinds of mistakes, with typical ones like 1) Regard vectors in V as columns of M. 2) Regard vectors in V just in R^4, without considering the constraint. 3) Check that matrix M is closed under addition and scalar multiplication. 4) Check specific M and vectors, instead of general ones. Basic Grading Policy: 1) Claim TRUE, and write the correct linear relationship between a,b and c,d, but mistakes on understanding of closedness of addition/scalar multiplication, 3-5pts. 2) Claim TRUE, and work toward showing the closedness, >=5pts, e.g. prove either closedness condition correctly is about 7pts. 3) Either of full answer as the two correct above is 10 pts. 4) In one exceptional case of claiming FALSE, with the wrong understanding that 'linear subspace' must be of dimensional one (line) but correctly pointing out V is of dimensional two (plane), >=7pts. Marco ----- True/False: Let A be nxn. Attaching a basis of NS(A) to one of CS(A) gives a basis of R^n. Answer: FALSE. Even though by the rank-nullity theorem the set consisting of the v's and w's together has the right number of vectors to be a basis, the set is usually not a basis, because it is usually a linearly dependent set. An example is given by the matrix (0 1 0 0) for which NS=CS=span{(1 0)}, i.e. v_1=w_1=(1 0). Clearly in this case the set {v_1 w_1} is not linearly independent, hence it is not a basis. Grading policy: 10 pts for valid counterexample and showing that the set {v_1,...,w_1,...} is not a basis. 8 pts for a correct counterexample with mistakes computing NS and CS. 6 pts for a matrix which is not a counterexample, but where you checked if the set {v_1,...,w_1,...} is a basis. 5 pts for incomplete reasonings to show that statement is false. 3 pts for restricting to a statement which is true and then trying - incorrectly - to show that it is false. For example: you deduce somehow that NS(M)=0. This is not the case in general, but when it happens, then the statement is true (for that very M). Trying to show that the statement for such M is false was assigned 3 points. A last note: the zero vectorspace has dimension zero, hence the set {0} consisting of one element is NOT a basis for it. Zach ---- The question was whether or not a basis for P_n must contain an element of degree k for each k between 0 and n inclusive. Answer: False. Most people offered counterexamples that were not a basis for Pn. Many people offered bases for subspaces of Pn. Others claimed that linear dependant sets or sets with fewer than n+1 elements were bases for Pn. These all received 3 points as the problem did say a basis B for Pn. Another common error was to claim that there are only n numbers between 0 and n. There are in fact n+1. This also only received the obligatory 3 points for correctly writing false. Obvious counterexamples were {x+1, x-1} in P2 and {X^n, x^n+x^(n-1, . . . , x^n+. . .+x+1} Any valid counterexample was given 10 points. Almost no partial credit was given. The only exception was if you stated that a basis for P_n could be written where all elements of the basis had degree n. Since no actual counterexample was given this received 5 points. Peter ----- True/False. The row space is perpendicular to the null space. Answer: TRUE. The correct solution was: If v is in NS(A), then Av=0. Write A as A= (r1 r2 r3 ... rn)^T (bad notation, but how else to type it?). Then 0 = Av = (r1 v, r2 v, ..., rn v). So the dot product of each row of A with v is 0. If u is in the row space of A, then u can be written as a linear combination of the rows of A: u=a1r1+a2r2+...+anrn. So, u \cdot v= (a1r1+ ... +anrn) \cdot v= a1(r1 \cdot v) + ... +an(rn \cdot v) =a1(0) + ... +an(0)=0. So if u is in RS(A) and v is in CS(A), then u \cdot v =0. Then most common mistake by far was to show that the dot product of each row of A with v was zero, but not to mention the fact that u in NOT one of these rows, but rather is in the span of the rows. Such a solution, if correct, got 6. Darren ------ If A is a square matrix such that (A-I)^2=0, then A is invertible. Answer: True. 0=(A-I)^2=(A-I)(A-I)=A^2-2A+I, so 2A-A^2=I=A(2I-A)=(2I-A)A, so 2I-A is the inverse of A. Point system according to answer given: True, good reason given= 10 points True, almost good reason (sign error)= 8 or 9 points True, good reason found amongst stranger arguments that don't work= 5 points True, no reason or bad reason= 3 or 4 points. False, yet bizarrely good reason for true written= 4 or 5 points False, no reason or bad reason= 0 points. Writing different True/False answer on the front than in the interior= 0 points Ben --- The grading scheme for problem 7 was as follows: 4 points each for correctly stating tight upper and lower bounds on the dimension, 5 points each for providing examples to show that the bounds are tight, 1 point each for proving that the numbers are upper and lower bounds. Some common errors: misunderstanding the definition of intersection failing to provide examples failing to prove that the numbers are upper and lower bounds asserting that subspaces are equivalent to matrices asserting that R^m is a subspace of R^n if m