Do the following vectors form a basis of , span , 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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- September 28th 2010, 09:56 AMcoolhandlukeVectors forming a basis,spanning
Do the following vectors form a basis of , span , 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? - September 28th 2010, 10:57 AMAckbeetQuote:

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

- September 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? - September 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?

- September 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? - September 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?

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

- September 28th 2010, 11:22 AMAckbeetQuote:

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

- September 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. - September 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 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 's. Make sense?

- September 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? - September 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?

- September 28th 2010, 12:59 PMcoolhandluke
Very much so. Thank you for your help.

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