# Spanning of a Set of Vectors

I'm studying for my finals and I came across the following question:

Consider the vectors: $V_1=(1,2,4),$ $V_2=(5,-1,3).$ Determine if these two vectors span $\mathbb{R}^3.$

I understand to do this you have to either find the determinant and if it equals zero it has no solutions, and thus the vectors do not span $\mathbb{R}^3$ and you can also see this if you use Gauss/ Gauss Jordan. I can't see a way of calculating the determinant using a $3\times 2$ system. So I have used Gauss, with the following row operations:

Showing that firstly $V_1=(1,2,4), V_2=(5,-1,3) =(a,b,c)$

$$\left( \begin{array}{ccc} 1 & 5 &| &a \\ 2 & -1 &| &b \\ 4 & 3 &| & c \end{array} \right)$$

$R_2 =2R_1 -R_2$

$$\left( \begin{array}{ccc} 1 & 5 & |&a \\ 0 & 9 & |&2a-b \\ 4 & 3 & |&c \end{array} \right)$$ $R_3=4R_1-R_3$ $$\left( \begin{array}{ccc} 1 & 5 & | & a \\ 0 & 9 & | & 2a-b \\ 0 & 17 & | & 4a-c \end{array} \right)$$

$R_2=R_2/9$

$$\left( \begin{array}{ccc} 1 & 5 & | & a \\ 0 & 1 & | & 2a/9 - b/9 \\ 0 & 17 & | & 4a-c \end{array} \right)$$ $R_3=-17R_2 +R_3$ $$\left( \begin{array}{ccc} 1 & 5 & | & a \\ 0 & 1 & | & 2a/9 - b/9 \\ 0 & 0 & | & 4a - c \end{array} \right)$$

I was wondering if this was the correct way of displaying although, i'm pretty sure my algebra is wrong...

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$\mathbb{R}^3$ has dimension 3. It cannot be spanned by two vectors in $\mathbb{R}^3$.

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I did think that, but this question was worth 7 marks, so that answer put me of some what, thanks though. – Xabi Apr 22 '12 at 17:00
Also, on your remark " I cant see a way of calculating the determinant using a 3X2 systems", this is because the determinant is only defined for square matrices. – nullUser Apr 22 '12 at 17:01
cool, i'll bear that in mind – Xabi Apr 22 '12 at 17:02

Keep in mind the following rule: if $\dim V =n$ and $\mathcal{B}$ is a set of $m<n$ vectors from $V$, then $\mathcal{B}$ cannot be a basis of $V$.

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The definition of dimension is the number of basis which span the vector space. If 2 vector can span the vector space, then it is at most 2 dimensions.

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