Seemingly impossible problem involving linear combination of vector components. Express $\langle 4, -8 \rangle$ as a linear combination of $\vec{u}$ and $\vec{v}$, given $\vec{u}=\langle 1,1 \rangle$ and $\vec{v}=\langle -1,1 \rangle$.
So, I set up: $\vec{i}=\langle 1,0 \rangle$ and $\vec{j}=\langle 0,1 \rangle \implies$
$$A(-\vec{i}+\vec{j})+B(\vec{i}+\vec{j})=4\vec{i}-8\vec{j}$$
This looks to be impossible to find values of $A$ and $B$ which satisfy this equality.
 A: Comparing the coefficients of $i$ and $j$ on both sides, you can get a set of two equations:
$$-A+B=4\\
A+B=-8$$
Is it possible now?
A: Therefore you have the simultaneous equations $$-A+B=4$$ $$A+B=-8$$
A: $$B\vec{u}+A\vec{v}=\langle B-A,B+A\rangle=\langle4,-8\rangle$$ leading
to:
$$B-A=4\wedge B+A=-8$$
Solution: $$B=-2\wedge A=-6$$
A: $\vec{u}=\langle 1,1 \rangle$
$\vec{v}=\langle -1,1 \rangle$
$A\vec{v} + B\vec{u} = \langle 4,-8 \rangle$
Note that
$\vec u \circ \vec v = 0$ and 
$\vec u \circ \vec u = \vec v \circ \vec v = 2.$
So
\begin{align}
   \vec u \circ (A\vec{v} + B\vec{u}) &= \vec u \circ\langle 4,-8 \rangle\\
   2B &= -4\\
   B &= -2
\end{align}
and 
\begin{align}
   \vec v \circ (A\vec{v} + B\vec{u}) &= \vec v \circ\langle 4,-8 \rangle\\
   2A &= -12\\
   A &= -6
\end{align}
A: $\vec u$ and $\vec v$ are two linearly independent vectors, in a vector space of dimension $2$: they form a basis, so there must be a solution.

By inspection,
$$-2\cdot\langle 1,1 \rangle-6\cdot\langle -1,1 \rangle=\langle 4,-8 \rangle.$$
