Recall that if we have m vectors u1, U2, ..., um in R", then we can form the matrix A whose columns are u1, ... , Um. Let B be the echelon form of A. Most questions have a yes no answer, but I am mostly interested in your reasons for the answer. Give full reasons for all answers. Suppose we are given the following matrix B:
[I'm gonna guess that B really looks like
(3 0 -1 5
0 0 2 -1
0 0 0 0)
and that 1 in 10000 is a typo (as is most of the rest, really).
(a) What is n?
There's actually no n mentioned in the question. I'll guess the vector space is supposed to be Rⁿ which means A and B are n×m matrices. So n is the length of each vector, the number of rows in A and B, which I guessed (because the matrix wasn't particularly formatted very well) was three.
Answer: 3
(b) What is m?
m vectors so m columns,
Answer: 4
(c) Are ui, ..., Um linearly independent?
We have a column of all zeros, second one. They can't be linearly independent with a column of zeros because there's always a non-trivial linear combination of the vectors that gives zero.
Answer: NO
(d) Does {U1, ... , um} span R ?
It obviously doesn't span R. The question probably should be does it span Rⁿ? All the vectors have a zero as their third element, so so will any linear combination of them. If we can only get 0 for the last element we can't be spanning the entire 3D space.
In echelon form to span the space we need a pivot in every row, meaning a leading non-zero term. There's none in the last row.
Answer: NO
(e) Looking at B can you write down a subset of the original set {U1, ..., Um} that would be guaranteed to be linearly independent?
We only have two non-zero rows so our subset has at most two vectors. We can choose any two of the three non-zero ones, how about
{ (-1,2,0)^T, (5,-1,0)^T }
These are vectors from B. The question is asking for vectors from A, which are what the u's are. We can't really work backwards to find the vectors from A but we know of the columns are independent from the echelon form they'll be independent in the original A as well.
Answer: { u₃, u₄ }
(f) Is there a subset of the original set {u1, ... , Um} that would be guaranteed to span R"?
Answer: NO
If the full set of vectors doesn't span Rⁿ there's no subset that will either.
(g) Write down a b e Rn for which Bx = b does not have a solution.
Bx is always going to have zero for that last coordinate, no matter what x is. So
Answer: b=(0,0,2)^T
is impossible. I wrote ^T because we're after a column vector, the transpose of the vector I typed.