1. May 29, 2014

### negation

1. The problem statement, all variables and given/known data

Is {(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3)} a spanning set for R3

3. The attempt at a solution

This is supposed to be easy but the answer sheet might be wrong.

The answer I have says it is and then proceed to say that (0;0;2) is linearly independent. But it isn't because (0;0;2) can be expressed as a linear combination of the other vectors!

2. May 29, 2014

### HallsofIvy

Staff Emeritus
Surely it doesn't say "(0; 0; 2) is linearly independent". For one thing that doesn't make sense. A single vector is not "dependent" or "independent". It must be dependent or independent of other vectors. You can also say that a set or vectors is "independent" (no vector is dependent on any of the others in the set) or "dependent" (at least one of the in the set is dependent on others).

In any case, you could answer the question directly from the definition: show that, for any numbers, x, y, and z, there are numbers a, b, c, and d, such that, a(1; 1; 0)+ b(0;0;2)+ c(0;0;1)+ d(1;2;3)= (a+ d; a+ 2; 2b+ c+ 3d)= (x, y, z). That is, show that the set of equations, a+ d= x, a+ 2b= y, 2b+ c+ 3d= z can be solved for any values of x, y, and z.

It is true in any vector space of dimension n, that a set of n independent vectors must span the space. So it is sufficient to find a subset of 3 independent vectors to show that these span the space.
For example, a(1;1;0)+ b(0;0;2)+ c(0;0;1)= (a; a+ b; c)= (0, 0 ,0) then we must have a= 0, a+ b= 0, and c= 0. Since b= 0, a+ b= a= 0. a= b= c= 0 is the only solution so that subset is independent so spans the space so the original set spans the space.
{(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3
{(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3

3. May 29, 2014

### negation

With respect to the question, the solution set has 1 free variable. Thus, the solution set has infinitely many solutions. And so, span {(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3)} is a spanning set for R4.

Keeping this in mind, I could have equally, instead of expressing [a;b;c;d] as a linear combination of the set of vectors, any vectors in R4 would have equally worked, wouldn't it?

I have also read that row reducing the set of vector
[1 1 1; 0 0 2; 0 0 1; 1 2 3] gives us the rank, which, if rank(matrix)= n for Rn then the solution set is trivial and so the system is linearly independent. But why?

4. May 29, 2014

### HallsofIvy

Staff Emeritus
?? You gave the problem as "Is {(1; 1; 0);(0; 0; 2);(0; 0; 1);(1; 2; 3)} a spanning set for R3?"

That is a "yes" or "no" question. I have no idea what you mean by a "solution set" for a "yes" or "no" question.