# Math Help - Vectors forming a basis,spanning

1. ## Vectors forming a basis,spanning

Do the following vectors form a basis of $R^3$, span $R^3$, or neither?

a1 = (1, 2, 1), a2 = (-1, 0, -1), a3 = (0, 0, 1)

Do I just check to see if they're linearly independent?

2. Do I just check to see if they're linearly independent?
Correct. Since the dimension of the space is 3, and you have 3 vectors, it follows that being a basis is equivalent to spanning the space.

3. Ok, so can I say this is true:

If the number of vectors equals the n in R^n, then they are spanning? So, in this case, if there are two vectors, they do not span? What if I have more than 3 vectors?

I can also say that if they span AND they are linearly independent, then the vectors form a basis?

4. If the number of vectors equals the n in R^n, then they are spanning?
Only if they are linearly independent. In that case, they are also a basis.

What if I have more than 3 vectors?
In three-dimensional space, more than 3 vectors could span the space, but they definitely won't be a basis, because they won't be linearly independent.

I can also say that if they span AND they are linearly independent, then the vectors form a basis?
Yes.

5. Thank you for your quick help.

For clarification:

If I have two vectors, for a 3-dimensional space, these vectors can not span or be a basis. This is true?

If I have four vectors, for a 3-d space, these vectors CAN span, but can not be a basis, since they will not be linearly independent. This is true?

How would you show that the 4 vectors span?

6. If I have two vectors, for a 3-dimensional space, these vectors can not span or be a basis. This is true?
True.

If I have four vectors, for a 3-d space, these vectors CAN span, but can not be a basis, since they will not be linearly independent. This is true?
True.

How would you show that the 4 vectors span?
Write an arbitrary vector $x$ in the space as a linear combination of the 4 vectors. That will generate a linear system of equations. If there are any solutions, then the vectors span. If there are not solutions, the vectors do not span. Make sense?

7. With the linear system of equations, what is the right hand equal to for the equations?

8. With the linear system of equations, what is the right hand equal to for the equations?
The components of your arbitrary vector $x$.

9. Hmm, I don't quite get it. I have four vectors: (-1, 2, 3), (0, 1, 0), (1, 2, 3), (-3, 2, 4).

Would I be solving for:

-1*X_1 + X_3 - 3*X_4 = Y_1
2*X_1 + X_2 + 2*X_3 + 2*X_4 = Y_2
3*X_1 + 3*X_3 + 4*X_4 = Y_3

I'm sure I didn't set that up right. There are 7 unknowns and 3 equations.

10. Your system is correct, but you're interpreting it incorrectly. In the context of solving that system, you don't treat your $Y_{j}$ as unknowns for which to solve. They are just arbitrary numbers. Treat them like you would treat the RHS of any system. You need to solve for the $X_{k}$'s. Make sense?

11. I can't solve this to get each X in terms of only Y's.

Does this mean it does not span?

12. You can't solve it exactly to get a unique solution. However, you're not after a unique solution (which, incidentally, would correspond with being a basis). You're after any solutions at all. Since the system is under-determined, if you have any solutions, you're going to have infinitely many solutions. You're checking to see whether you have zero solutions, or infinitely many solutions. Zero solutions means you don't have a spanning set. Infinitely many solutions means you have a spanning set, but it's not a basis. Exactly one solution means you have yourself a basis. Make sense?

13. Very much so. Thank you for your help.

14. Great. You're welcome. Have a good one!