Question

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: (3 0 -1 5 0 0 2 -1 10000)
(a) What is n?

(b) What is m?

(c) Are ui, ..., Um linearly independent?

(d) Does {U1, ... , um} span R ?

(e) Looking at B can you write down a subset of the original set {U1, ..., Um} that would be guaranteed to be linearly independent?

(f) Is there a subset of the original set {u1, ... , Um} that would be guaranteed to span R"?

(g) Write down a b e Rn for which Bx = b does not have a solution.

(h) Write down a b e R" for which Bx = b has a unique solution.

(i) Is there a new vector w E Rthat you could add to the set {u1, ..., Um} to guarantee that {U1, ... , um, w} will span R ?

(i) Is there a column of B that is in the span of the rest? If so, find it (k) Put B into reduced echelon form.

Write down a non-zero solution of Ax = 0 if you can. (m) How many free variables are there in the set of solutions to Ax = b when there is a solution? (n) If you erased the last row of zeros in B then would the columns of the resulting matrix be linearly independent?

Can you add rows to B to make the columns of the new matrix linearly independent? If yes, give an example of the new matrix you would construct.

Answer 1

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.