# Finding a vector not included in this span

1. Dec 15, 2012

### lo2

1. The problem statement, all variables and given/known data

I have this span, spanned by these three vectors in R^5:

$\underline{a_1}= \left( \begin{array}{c} 2 \\ 3 \\ 1 \\ 4 \\ 0 \end{array} \right)$

$\underline{a_2}= \left( \begin{array}{c} 1 \\ -1 \\ 2 \\ 4 \\ 3 \end{array} \right)$

$\underline{a_3}= \left( \begin{array}{c} 3 \\ 4 \\ -1 \\ 3 \\ 5 \end{array} \right)$

2. Relevant equations

3. The attempt at a solution

Well I thinking about looking at this equation (where A consists of a1, a2 and a3):

Ax=b

And then reduce A to an identity matrix, and where b is just any vector, b = (b1, b2, b3, b4, b5). And then I could decide b so that it is not a solution to this equation, which means it cannot be written as a linear combination of the three a's, which means it is not in the span of these vectors.

So is the correct approach?

And Maple does not seem to want to solve this matrix with all these unknowns, so have you any idea why that is?

2. Dec 15, 2012

### Zondrina

So wait, you have the span of those three vectors and you're trying to determine if you can make the set of vectors smaller?

If so, solve Ax = 0, not Ax = b. I might be misunderstanding you here.

Last edited: Dec 15, 2012
3. Dec 16, 2012

### lo2

No I just have to come up with a vector in R^5 that does not lie in span(a1, a2, a3).

4. Dec 16, 2012

### lurflurf

Try the row reduction again. Also express in this basis the vector

$$\underline{a_4}= \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 4 \\ 0 \end{array} \right)$$

5. Dec 16, 2012

### lo2

Just came to think of something:

I have expanded a1, a2, and a3 to a basis for R^5 by adding e1 = (1,0,0,0,0) and e2 = (0,1,0,0,0).

So since they are in this basis they have to be linear independent of a1, a2 and a3 right? And hence e1 and e2 do not lie in the span of a1, a2 and a3?

6. Dec 16, 2012

### lurflurf

Yes you can pick two vectors, just check they are not in the span of previous vectors. Those two are not.

7. Dec 16, 2012

### lo2

Ok thanks for the help! :)