problem 1 exam 3

1a. Assume A is nxn. Otherwise, there is no chance to get such an
    equivalence. No, it was not assumed, a mistake: you must add the
    assumption to the problem statement. Observe that det(B)det(A) not
    zero implies det(A) not zero, without regard the value of det(B).

1b. In part (2), you must prove that k<n using vector space theorems.
    Given was A nxn, which is where you get the value of n. The number
    k is given by properties in the hypothesis. Yes, it is easy to
    prove that A^n = 0, once you know that k<n.

1c. Some people wanted to use determinants on this problem. That won't
    work because A,B are not assumed square.

1d. A-2I = P^(-1) (B - 2I) P is given. Expand. See if a similarity
    relation emerges for A,B.


problem 2 exam 3

2a. Ex 5 p 150 discusses W, but fails to prove closure of plus and dot.
    Also, the 8 properties of a vector space are not proved ther - you
    are to supply all proofs. Expect short, concise one-line proofs.

2b. Like the valet lot analogy, W is the big lot and V is the valet lot.
    The set V arises by adding restrictions onto objects in W. Settle
    the subspace question using the subspace criterion: V is not empty
    and cv1+c2v2 is in V for constants c1,c2 and vectors v1,v2 in V.

2c. You are to show T(ax+by)=aT(x)+bT(y). Then find a formula for
    the objects in ker(T), that is, describe all sequences x such 
    that T(x)=zero sequence.

2d. Let S be defined by S(f)=2f-f' on the set of a functions f
    which have domain 0 <= t <= 1 and for which f,f',f'', ...
    all exist and are continuous (space of infinitely differentiable
    functions on [0,1]). Solve 2f-f'=0 to find ker(S) and then
    nullity(S) is the dimension of ker(S).


problem 3 exam 3.

3a. Square both sides of the inequality to obtain an equivalent one
<u+v,u+v> <= <u,u>+2<u,v>+<v,v>. Then use CSB.

3b. Let u1,u2,u3 be the given vectors. Use Gram-Schmidt to find
orthogonal v1,v2,v3 with the same span, each of length 1. Now use
the projection formula <v,v1>v1+<v,v2>v2+<v,v3>v3.

3c. Let u1,u2,u3 be the columns of A. Perform Gram-Schmidt on them
to obtain orthogonal vectors v1,v2,v3. Let w1,w2,w3 be the unit vectors
constructed from v1,v2,v3. Let Q=aug(w1,w2,w3). Let R be constructed
as in Otto's book page 198, so that r11 = length of v1, etc. Write out
Q and R in detail, then check in maple that A=QR.

3d. Suppose A has an inverse B, AB=BA=I, and A=QR for Q=Q1,Q2 and 
R=R1,R2. Show Q1=Q2 and R1=R2. See problem 30 section 5.3.

problem 4 exam 3.

4a. Which is the easiest?

4b. The relation T=E1 E2 E3 ... A holds for some elementary matrices 
E1,E2,E3,..; this relation plus the product theorem for determinants
gives the result.

4c. Mystery problem. The matrix of all ones is not invertible, so there 
has to be at least one zero in A. The answer depends on the dimension, 
perhaps, so experiment with several dimensions before reporting your 
answer. Supply a proof of your statement, meaning that fewer than the 
claimed number of zeros causes A to be non-invertible.

4d. You will need to know that rref(aug(A,I))=aug(I,B) iff AB=BA=I.
Then rref methods on C=aug(A,I) compute the inverse B as the last 
columns of rref(C). For the adjoint method, see page 281.