# How can I find two vectors in a given $\operatorname{span} \{u, v\}$ that are not multiples of $u$ or $v$?

But do appear to be linear combinations? $u$ and $v$ are $3$-component vectors. The question posed is:

Find two vectors in $\operatorname{span}\{u,v\}$ that are not multiples of $u$ or $v$ and show the weights on $u$ and $v$ used to generate them.

I.e. I am looking for two vectors $au + bv$ and $cu + dv$, and I take it from the wording of the question that the one restriction on these coefficients is that they are all non-zero (or I'd just have a multiple of $u$ or $v$).

Some things I understand or have already demonstrated to myself:

• $u$ and $v$ are linearly independent
• together they describe a plane in $3$-dimensional space
• the problem setter wants vectors that are 'off' the directions described by $u$ and $v$ (i.e not parallel).

From this, I thought any two linear combinations would be fine, but none of the ones I've submitted are acceptable. Is there something in this question that I have misunderstood? I would really appreciate any help understanding what is being asked (rather than being given the coefficients).

-
Hint: draw a picture of (generic) $u$ and $v$ in the $2$-dimensional plane spanned by them, and from there try to guess "easy" values for $a, b, c, d$ (start with $\pm 1$). – student Feb 26 '12 at 16:59
Your reasoning sounds fine, so it's odd that your answers aren't working. Could you tell us what the vectors $u$ and $v$ are, and what you've submitted? Maybe things go wrong when translating between the vector as a $3$-tuple (in the standard basis) and as a weighted sum of $u$ and $v$? – Dylan Moreland Feb 26 '12 at 17:03
@Dylan: $u$=<-3, 1, -3> and $v$=<3, -4, 1>. Right now I just tried plugging in 1 and 2 as the coefficients of $u$ and $v$ respectively to achieve the first new vector, and 2 and 1 for the second one. That's all I can fill in on the LHS - coefficients only. The RHS asks for the vector each linear combination makes. I've filled them in as <3,-7,-1> and <-3,-2,-5> (I really hope this doesn't turn out to be an embarrassing lack of basic arithmetic!) The automatic marker returns correct or incorrect for each of the four coefficients and the two new vectors. I got them all wrong. Does this help? – Katherine Rix Feb 26 '12 at 17:32
@Leandro: I started with the picture as suggested :) However...I don't know what I'm guessing at! What am I trying to achieve? – Katherine Rix Feb 26 '12 at 17:34
if $u,v$ are linearly independent, $u+v,u-v$ always works. perhaps the software will only recognize this as the "correct" answer (even though there are many others). – David Wheeler Mar 6 '12 at 1:38