# Do four Vectors form a Basis in R3?

If I have 4 Vectors: $a_1 = (-1,2,3), a_2 = (0,1,0), a_3 = (1,2,3), a_4 = (-3,2,4)$ How can I determine if they form a basis in R3?

I set the Matrix up into a 3X4 matrix and then reduced it down to the identity matrix with an additional vector $(13/6,-2/3,-5/6)$.

I also know that for it to form a basis it needs to be linear independent which implies $c1*w1+c2*w2+c3*w3+c4*w4=0$ . If I calculated expression where $c_1=(-x+z-3x), c_2=(y-2x-4/6(z-3x)), c_3=(z-3x)$ and since we want to show $x=y=z=0$, would that mean that these four vectors would NOT form a basis but because there is a fourth vector within the system therefore it is inconsistent? Therefore not providing a Span for R3 as well?

I think I have the math and the concepts down. I would like for someone to verify my logic for solving this and help me develop a proof.

• There's a lot wrong with your third paragraph and it's hard to know where to start. At the very least: the vectors do span $\Bbb R^3$. It is not clear what has led you to believe otherwise. Commented Feb 2, 2017 at 15:42
• I was using the row transformations to map out what the Scalar constants where. But more importantly my questioned pertained to the 4th vector being thrown out. Commented Feb 2, 2017 at 15:48
• Spanning a space and being linearly independent are separate things that you have to test for. No set of $\gt3$ vectors in $\mathbb R^3$ can be linearly independent (see the definition of dimension), but whether or not that set spans all of $\mathbb R^3$ depends on how many independent vectors you do have, which you can’t determine without probing further.
– amd
Commented Feb 2, 2017 at 17:15

No. We have a theorem:

Basis Theorem. Let $V$ be a vector space of dimension $n$. Then any basis of $V$ will contain exactly $n$ linearly independent vectors.

Since your set in question has four vectors but you're working in $\mathbb{R}^3$, those four cannot create a basis for this space (it has dimension three).

Now, any linearly dependent set can be reduced to a linearly independent set (and if you're lucky, a basis) by row reduction. Check for unit vectors in the columns - where the pivots are. Any column that is not a unit vector (a vector with a $1$ in exactly one position, zeros everywhere else) corresponds to a vector that can be thrown out of your set.

More concretely, let $S = \{ (-1, 2, 3)^T, (0, 1, 0)^T, (1, 2, 3)^T, (-3, 2, 4)^T \}.$ As you said, row reductions yields a matrix

$$\tilde{A} = \begin{pmatrix} 1 & 0 & 0 & 13/6 \\ 0 & 1 & 0 & -2/3\\ 0 & 0 & 1 & -5/6 \end{pmatrix}$$

in which each column corresponds to the proper vector in $S$ (first column corresponds to the first vector, ...). The last column does not have a pivot, and so the last vector in $S$ can be thrown out of the set. The remaining members of $S$ not only form a linearly independent set, but they span $\mathbb{R}^3$, and since there are exactly three vectors here and $\dim \mathbb{R}^3 = 3$, we have a basis for $\mathbb{R}^3$.