Does $\{(1, -1),(2, 1)\}$ spans $\mathbb{R}^2$? please correct me Can anyone please correct me? my problem is in the proof part below
Q: Does $\{(1, -1),(2, 1)\}$ spans $\mathbb{R}^2$?
A:
$$c_1(1, -1) + c_2(2, 1) = (x, y)$$
$$c_1 + 2c_2 = x$$
$$-c_1 + c_2 = y$$

$$c_1 = x - 2c_2$$
$$-(x - 2c_2) + c_2 = y$$
$$-x + 2c_2 + c_2 = y$$
$$c_2 = \frac{x + y}{3}$$

$$c_1 + 2\frac{x + y}{3} = x$$
$$c_1 = x - 2\frac{x + y}{3}$$
$$c_1 = x - \frac{2x-2y}{3}$$
Conclusion: it can reach any point $(x, y)$ in $\mathbb{R}^2$
Proof: Get to $(4, 7)$:
so, $c_1 = 6$, $c_2 =\frac{11}{3}$.
$$6(1, -1)+\frac{11}{3}(2, 1) = \left(\frac{40}{3}, -\frac{7}{3}\right)$$
!!! it should've been equal to $(4, 7)$

 A: You can do it much simpler! You know that $(1,0)$ and $(0,1)$ span $\mathbb{R}^2$. Thus if you can describe those two vectors using $(1,-1)$ and $(2,1)$, these two vectors span the whole space $\mathbb{R}^2$ as well:
\begin{align}
(1,0) &= \frac{1}{3}\cdot(1,-1) + \frac{1}{3}\cdot(2,1) \\
(0,1) &= 2\cdot (1,-1) +(-1)\cdot (2,1)
\end{align}
A: Your idea is good, let's try rewriting it in a clearer way. You want to show that, for every $(x,y)$, you can find $c_1$ and $c_2$ such that
$$
c_1(1,-1)+c_2(2,1)=(x,y)
$$
that translates into the linear system
$$
\begin{cases}
c_1+2c_2=x\\
-c_1+c_2=y
\end{cases}
$$
Summing up the two equations gives $3c_2=x+y$, so
$$
c_2=\frac{x+y}{3}
$$
The second equation becomes $c_1=c_2-y$, that is
$$
c_1=\frac{x+y}{3}-y=\frac{x-2y}{3}
$$
In the particular case of $(x,y)=(4,7)$ we get
$$
c_1=\frac{4-14}{3}=-\frac{10}{3},\qquad c_2=\frac{4+7}{3}=\frac{11}{3}
$$
Indeed,
$$
-\frac{10}{3}(1,-1)+\frac{11}{3}(2,1)=(4,7)
$$
Some more care in your computations would have spared you with the errors.
A: Perhaps easier:
$$\begin{cases}&\;\;\;c_1+2c_2=x\\{}\\&-c_1+c_2\;\;=y\end{cases}\stackrel{\text{sum both eq's}}\implies3c_2=x+y\implies c_2=\frac{x+y}3$$
and substituting back, say in equation two:
$$c_1=\frac{x-2y}3$$
A: Q. Does $\{(1,−1),(2,1)\}$ spans $\Bbb{R}^2?$
Answer :
Yes
$\begin{vmatrix}1&2\\-1&1\end{vmatrix}=3\neq0$
Implies $A=\begin{pmatrix}1&2\\-1&1\end{pmatrix}$ is invertible and hence $Ax=b$ has a solution (in fact unique) for every $b\in \Bbb{R^2}$
Thus every vector of $\Bbb{R}^2$ is a linear combination of the columns of $A$.
So in that case , only you need to show determinant in non zero.
A: The space $\mathbb{R}^2$ has dimension 2, so it is spanned by any two linearly independent vectors. So, you just need to check that $(1,-1)$ and $(2,1)$ are linearly independent, i.e. to check that if $c_1(1,-1) + c_2(2,1) = (0,0)$ you must have $c_1=c_2=0$.
$$
\begin{cases}
c_1+2c_2 = 0\\-c1+c2=0
\end{cases}\Leftrightarrow
\begin{cases}
3c_2 = 0\\ c_1=c_2
\end{cases}\Leftrightarrow
\begin{cases}
c_1 = 0\\ c_2=0
\end{cases}
$$
This is equivalent to checking that the determinant of a matrix that contains $(1,-1)$ and $(2,1)$ in rows (or columns) is invertible.
$$
