# Determine whether the given vectors form a basis

1. May 12, 2012

### robertjford80

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

w1 = 2 1 2
w2 = 1 -2 -3
w3 = 5 0 1

for
R3

2. Relevant equations

3. The attempt at a solution

The books says the above is not a basis, why not? There are no free variables, none of the vectors are multiples of the other, they are linearly independent and the number of unknowns equals the number of equations. That check list to me signals a basis

2. May 12, 2012

### chrisb93

What makes you think they are lineally independent?

3. May 12, 2012

### robertjford80

As I already said: There are no free variables, none of the vectors are multiples of the other

This is what I get in Reduced Ech form

1 -2 -3
0 5 -4
0 0 -6

4. May 12, 2012

### Quinzio

Actually is:
$$w_3=2w_1+w_2$$

5. May 12, 2012

### chrisb93

In echelon form I get
$\left( \begin{matrix} 1 & -2 & -3 \\ 0 & 5 & 8 \\ 0 & 0 & 0 \end{matrix} \right)$

Also be careful with difference between echelon form and reduced echelon form. The matrix you wrote (and the one above) is in echelon form, as every pivot has only zeros below it. Whereas in reduced echelon form the piviot is the only non zero entry in the column i.e. the above matrix becomes:

$\left( \begin{matrix} 1 & 0 & \frac{1}{5} \\ 0 & 1 & \frac{8}{5} \\ 0 & 0 & 0 \end{matrix} \right)$

6. May 12, 2012

### robertjford80

Ok, I understand now.

Also thanks for the tip about echelon form and reduced echelon form, I thought they were the same.