Determine whether the set spans $R^2$
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!
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$\begingroup$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 $c_1,c_2$.
You started correct, but remember that your goal is to prove/disprove linear independence of at least $2$ (And that's enough here) of the vectors. So you would want to put $(0,0)$ at the right hand side of your equation (Which is really a homogeneous system).
With $2$ vectors, this is the mathematical formulation of asking "Do these $2$ vectors share the same direction? i.e, Is one a scalar multiple of the other"?
If the answer is yes (i.e a non-trivial solution is found), you can combine them all you want; You are never never going to produce a vector with a different direction.
If the answer is no (i.e only the trivial solution satisfies the equation), they do not share the same direction, and you can express any vector using some combination, just try.
Once you know $2$ of the $3$ vectors are independent, then they form a basis for $\mathbb{R}^2$ by definition. Every basis spans It's space, and adding a "third wheel" makes no difference with respect to the spanning property.
The case for more than $2$ vectors in a combination is principally the same.
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