Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

Thanks in advance!

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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

share|improve this answer
    
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 $V$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.