Question

Determine whether the given set of vector spans R^2? {(1,-1), (2,-2), (2,3)}
Please, explain

Answer 1

let u = (1,-1)
v=(2,-2)
t =(2,3)
I have u = 2v, the given vectors linearly dependent .

Answer 2

so, can I get rid of v? just consider u and t?

Answer 3

@ChristopherToni

Answer 4

sorry v = 2u

Answer 5

The first thing to note (which you already did) is that (1,-1) and (2,-2) are multiples of one another; in particular, (2,-2) = 2(1,-1). Thus, the set of vectors \(\{(1,-1),(2,-2),(2,3)\}\) are not linearly independent; however, that doesn't necessarily rule out the possibility of it spanning all of \(\large \Bbb{R}^2\). Because of the fact that (2,-2) = 2(1,-1), it turns out that \(\{(1,-1),(2,-2),(2,3)\}\) and \(\{(1,-1),(2,3)\}\) generate the SAME spanning set. At this point, you need to ask yourself if (1,-1) and (2,3) are linearly independent. If you can show they are linearly independent, then the set \(\{(1,-1),(2,-2),(2,3)\}\) (and thus \(\{(1,-1),(2,3)\}\)) spans \(\large \Bbb{R}^2\).

A completely different way of determining if this set of vectors spans \(\Bbb{R}^2\) is to write the vectors in one matrix
\[\begin{bmatrix}1 & -1\\ 2 & -2\\ 2 & 3\end{bmatrix}\]and then get it into row echelon form. If the rank of the resulting matrix is 2, then the set spans \(\large \Bbb{R}^2\), and if the rank is 1, then the set spans \(\large \Bbb{R}\).

Does this make sense? :-)

Answer 6

Thanks for clearly explanation. I understood. I review for next Monday final. After the course, surely we know how to solve the problem in easiest way. However, my prof always says: "use this method to solve" or " must stick with that method" That's why I confused and wonder why don't we use the easiest one to solve. hehehe

Answer 7

can I ask one more? same topic?

Answer 8

Sure. :-)

Answer 9

why {(6,-2), (-2,2/3), (3,-1) } doesn't span R^2?

Answer 10

Well it's obvious that \(\large (6,-2) = 2(3,-1)\). However, the tricky one is seeing that \(\large (-2,\frac{2}{3}) = -\frac{2}{3}(3,-1)\). XD

Therefore, each vector is a multiple of \(\large (3,-1)\) and thus the set \(\large \{(6,-2),(2,-\frac{2}{3}),(3,-1)\}\) spans \(\large\Bbb{R}\), not \(\large \Bbb{R}^2\).

Answer 11

WAAAAT? why is it easier to you but not to me like that.?? I am jealous , man!!

Answer 12

Thanks for the help, genius!!

Answer 13

Trust me, it wasn't that easy at first for me either. I've been doing linear algebra for over 5 years now, so I gained quite a lot of insight over that period of time. You get better at seeing these kind of things when you do thousands upon thousands of problems like this. XD