# Thread: Determine if each of the following sets is a basis for the indicated subspace

1. ## Determine if each of the following sets is a basis for the indicated subspace

{[1,2,3]T,[4,5,6]T,[7,8,9]T} in R3

So, I've learned that for a set of vectors to be a basis it has to satisfy 2 of 3 of the following rules (if it meets two, the third is given):
1.) Must be linearly independent
2.) Must be a spanning set
3.) the set has dimension(V) elements

I've shown that the set is not linearly independent so it doesn't meet #1.
The set does meet #3.

I am running in to some confusion on how to test if it is a spanning set. Would I set up an augmented matrix and solve to test for consistency?
Also, I had previously thought that if a set was LI, then it could not be a basis...

2. ## Re: Determine if each of the following sets is a basis for the indicated subspace

$\mathbb{R}^3 \text{ has 3 dimensions so we must have 3 linearly independent vectors of 3 elements each in order to span it.}$

$\text{Since the 3 vectors are not linearly independent they cannot be a spanning set of }\mathbb{R}^3$

A more interesting problem is when you have greater than 3, 3 dimensional vectors.

Clearly they aren't all linearly independent but do they span $\mathbb{R}^3?$

They do if a subset of 3 of them are linearly independent.

3. ## Re: Determine if each of the following sets is a basis for the indicated subspace

Ok, then being that they are not linearly independent, and not a spanning set then it is not a basis?

Also, I noticed that the middle vector is the average of the first two vectors.. does this mean I can remove that and get a basis?

4. ## Re: Determine if each of the following sets is a basis for the indicated subspace

Originally Posted by MrJank
Ok, then being that they are not linearly independent, and not a spanning set then it is not a basis?

Also, I noticed that the middle vector is the average of the first two vectors.. does this mean I can remove that and get a basis?
Yes, a basis both spans it's space and it's elements are linearly independent.

yes but only for a 2 dimensional subspace of $\mathbb{R}^3$, i.e. a plane