Determine whether the given vectors form a basis

[robertjford80]

Homework Statement

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

for
R³

Homework Equations

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

 

[chrisb93]

What makes you think they are lineally independent?

 

[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

 

[Quinzio]

Actually is:
[tex]w_3=2w_1+w_2[/tex]

 

[chrisb93]

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

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:

[itex]\left( \begin{matrix}
1 & 0 & \frac{1}{5} \\
0 & 1 & \frac{8}{5} \\
0 & 0 & 0
\end{matrix} \right)[/itex]

 

[robertjford80]

Ok, I understand now.

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