[Math] Determine whether the set spans $R^2$

linear algebravector-spaces

So I'm told to determine whether a set spans $R^2$ and if it doesn't then give a geometric description of the subspace that it does span.

$S=\{\left(-1,4\right),\left(4,-1\right),\left(1,1\right)\}$

I first tried to form a linear combination:

$c_1(-1,4)+c_2(4,-1)+c_3(1,1)=(u_1,u_2)$

$(c_1+4c_2+c_3, 4c_1-c_2+c_3)=(u_1,u_2)$

Which gives the system:

$-c_1+4c_2+c_3=u_1$

$4c_1-c_2+c_3=u_2$

That system makes sense to me, but the solution manual says the below system is equivalent to the above system, but I don't know how they came up with the below system:

$c_1-4c_2-c_3=-u_1$

$15c_2-5c_3=4u_1+u_2$

In either case I understand how to get to this point but I don't really know how to solve the 'problem.' Any help in understanding this is greatly appreciated!

Best Answer

To span $\mathbb{R}^2$, two vectors suffice. Let $x=(4,-1)$ and $y=(1,1)$ (for instance): show $x$ and $y$ are independent, by showing that if $\alpha x+\beta y=0$ for some reals $\alpha,\beta$, then one must have $\alpha=\beta=0$.

As for the systems you consider (which is not the simplest way to prove $S$ spans $\mathbb{R}^2$):

  • the first is equivalent to the second, as the latter is just a rewrting of the former separating the two component equalities in two equations ($a,b)=(c,d)$ is equivalent to $a=c\text{ and }b=d$).
  • the second and the third are equivalent, as the one can get the other by linear combinations. The third is obtained by
    • multiplying the first equation of the second (2.1) by $-1$, to get the first equation (3.1);
    • adding $4$ times the first equation (2.1) to the second equation (2.2), to get its second equation (3.2) Now, you have a system with three unknowns ($c_1,c_2,c_3$) and two equations -- hence underdetermined. A simple way to solve it would be to set $c_3=0$ (for instance) and solve in $c_1,c_2$.
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