Finding a Basis for a set of vectors

[Smazmbazm]

Homework Statement

Let H be the set of all vectors of the form $(a-3b, b-a, a, b)$ where $a$ and $b$ are arbitrary real scalars. Show that H is a subspace of $ℝ^4$ and find a basis for it.

Right, I've shown it's a proper subspace, just need help with finding a basis. Is ${a-3b, b}$ a suitable basis for this?

I achieved this but putting the vectors into matrix form

1 -3
-1 1
1 0
0 1

Reducing them to end up with

1 -3
0 0
0 0
0 1

Is this a proper way to solve this question? Thanks in advanced.

 

[pdxautodidact]

I believe so. That basis spans a two dimensional substance, because a-3b, and b-a are both linear combination of the remaining two vectors: a, b. By reducing you show that the two were linear combinations, so they went to zero, and you're left with the two basis vectors. Consider one last modification, though:

$\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array} \right]$

The space is just span(a,b).

 

[Smazmbazm]

Great, thanks for that. Makes sense

 

[HallsofIvy]

What, exactly is your answer? Once you have row reduced, the only 4 dimensional vectors we can get from that matrix are (1, 0, 0, 0) and (-3, 0, 0, 1). That clearly cannot be a basis for this space because the second and third components would always be 0 which is not the case here.

And what do you mean by "Is a−3b,b a suitable basis for this?". a and b, and so a- 3b, are numbers, not vectors and so cannot be a "basis".

Here is how I would do it. H is the set of all vectors of the form (a- 3b, b- a, a, b).

(a- 3b, b- a, a, b)= (a, -a, a, 0)+ (-3b, b, 0, b)= a(1, -1, 1, 0)+ b(-3, 1, 0, 1).

And now a basis, and the dimension, are obvious.

 

[pdxautodidact]

It's been so long since I've had linear algebra, that is exactly how it's done. I got caught up with the linear dependence. But, HallsofIvy is exactly correct

$$
a \left(\begin{array}{c}1 \\ -1 \\ 1 \\ 0\end{array}\right) + b\left(\begin{array}-3 \\ 1 \\ 0 \\ 1\end{array}\right) $$gives you the basis vectors. While the subsspace is still dim = 2, it's formed by span$\{\left(\begin{array}{c}1 \\ -1 \\ 1 \\ 0\end{array}\right),\left(\begin{array}-3 \\ 1 \\ 0 \\ 1\end{array}\right)\}$.

Sorry about the dodgy reply.