Finding a Basis for a set of vectors

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  • #1
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Homework Statement



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

Right, I've shown it's a proper subspace, just need help with finding a basis. Is [itex]{a-3b, b}[/itex] 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.
 

Answers and Replies

  • #2
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).
 
  • #3
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Great, thanks for that. Makes sense
 
  • #4
HallsofIvy
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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.
 
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  • #5
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.
 

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