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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ben Grossmann
    Commented Feb 2, 2017 at 15:42
  • $\begingroup$ 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. $\endgroup$
    – User2648648
    Commented Feb 2, 2017 at 15:48
  • $\begingroup$ 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. $\endgroup$
    – amd
    Commented Feb 2, 2017 at 17:15

1 Answer 1

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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 $\mathbb{R}^3$.

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