Do the following vectors form a basis of $\displaystyle R^3$, span $\displaystyle R^3$, 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?

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- Sep 28th 2010, 09:56 AMcoolhandlukeVectors forming a basis,spanning
Do the following vectors form a basis of $\displaystyle R^3$, span $\displaystyle R^3$, 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, 10:57 AMAckbeetQuote:

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

- Sep 28th 2010, 11:05 AMcoolhandluke
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, 11:08 AMAckbeetQuote:

If the number of vectors equals the n in R^n, then they are spanning?

Quote:

What if I have more than 3 vectors?

Quote:

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

- Sep 28th 2010, 11:11 AMcoolhandluke
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, 11:18 AMAckbeetQuote:

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

Quote:

How would you show that the 4 vectors span?

- Sep 28th 2010, 11:21 AMcoolhandluke
With the linear system of equations, what is the right hand equal to for the equations?

- Sep 28th 2010, 11:22 AMAckbeetQuote:

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

- Sep 28th 2010, 11:28 AMcoolhandluke
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, 11:34 AMAckbeet
Your system is correct, but you're interpreting it incorrectly. In the context of solving that system, you don't treat your $\displaystyle Y_{j}$ 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 $\displaystyle X_{k}$'s. Make sense?

- Sep 28th 2010, 11:42 AMcoolhandluke
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, 11:47 AMAckbeet
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, 12:59 PMcoolhandluke
Very much so. Thank you for your help.

- Sep 28th 2010, 01:00 PMAckbeet
Great. You're welcome. Have a good one!