# Vectors forming a basis,spanning

• Sep 28th 2010, 10:56 AM
coolhandluke
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?
• Sep 28th 2010, 11:57 AM
Ackbeet
Quote:

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.
• Sep 28th 2010, 12:05 PM
coolhandluke
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?
• Sep 28th 2010, 12:08 PM
Ackbeet
Quote:

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.

Quote:

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.

Quote:

I can also say that if they span AND they are linearly independent, then the vectors form a basis?
Yes.
• Sep 28th 2010, 12:11 PM
coolhandluke
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?
• Sep 28th 2010, 12:18 PM
Ackbeet
Quote:

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

Quote:

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.

Quote:

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?
• Sep 28th 2010, 12:21 PM
coolhandluke
With the linear system of equations, what is the right hand equal to for the equations?
• Sep 28th 2010, 12:22 PM
Ackbeet
Quote:

With the linear system of equations, what is the right hand equal to for the equations?
The components of your arbitrary vector $x$.
• Sep 28th 2010, 12:28 PM
coolhandluke
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.
• Sep 28th 2010, 12:34 PM
Ackbeet
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?
• Sep 28th 2010, 12:42 PM
coolhandluke
I can't solve this to get each X in terms of only Y's.

Does this mean it does not span?
• Sep 28th 2010, 12:47 PM
Ackbeet
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?
• Sep 28th 2010, 01:59 PM
coolhandluke
Very much so. Thank you for your help.
• Sep 28th 2010, 02:00 PM
Ackbeet
Great. You're welcome. Have a good one!