Write a linear system in a new basis thanks for spending some time helping :)
How exactly do you represent a linear system in a different basis?
Let's say the new basis B = {(1,1),(-1,1)}
How do I write the system
\begin{cases} x+y=7\\ 2x–4y=-16 \end{cases}
in the new basis?
 A: In terms of matrices your system is
$$\left[\begin{array}{cc}1&1\\2&-4\end{array}\right]
\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}7\\-16\end{array}\right],$$
but if your basis change is
$$v_1=e_1+e_2,$$
$$v_2=-e_1+e_2,$$
then the old basis in terms of the new one is expressed as
$$e_1=\frac{1}{2}v_1-\frac{1}{2}v_2,$$
$$e_2=\frac{1}{2}v_1+\frac{1}{2}v_2.$$
The basis change associated matrix is
$B=\left[\begin{array}{cc}1&-1\\1&1\end{array}\right]$.
If the problem above can be written as
$Tu=w$
then by multiplying with $B^{-1}$ you will get the equivalent problem given by
$$(B^{-1}TB)B^{-1}u=B^{-1}w,$$
which is
$$\left[\begin{array}{cc}0&-3\\-2&-3\end{array}\right]
\left[\begin{array}{c}\bar x\\\bar y\end{array}\right]=
\left[\begin{array}{c}-\frac{9}{2}\\-\frac{23}{2}\end{array}\right],$$
where the new variables are
$$\bar x=\frac{1}{2}x+\frac{1}{2}y,$$
$$\bar y=-\frac{1}{2}x+\frac{1}{2}y.$$
So the system is
\begin{cases} 3\bar y=\frac{9}{2}\\ \\2\bar x+3\bar y=\frac{23}{2}\end{cases}
Addendum
The original problem's solution is
$u_0=\left[\begin{array}{c}2\\5\end{array}\right]$
and the transformed problem has the solution
$\left[\begin{array}{c}\frac{7}{2}\\\frac{3}{2}\end{array}\right]$, that perfectly coincides with $B^{-1}u_0$.
A: Any matrix multiplication is a projection over new basis, if you have:
$$\left[\begin{array}{cc}1&1\\2&-4\end{array}\right]
\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}7\\-16\end{array}\right],$$
and you want write it in new basis:
$$e_1=\begin{bmatrix}1\\1\end{bmatrix}, e_2=\begin{bmatrix}-1\\1\end{bmatrix}$$
then multiply internally by new basis.
$$\begin{bmatrix}1&1\\2&-4\end{bmatrix}\begin{bmatrix}1&-1\\1&1\end{bmatrix}^{-1}(\begin{bmatrix}1&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix})=\begin{bmatrix}7\\-16\end{bmatrix}$$
where,
$$\begin{bmatrix}1&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\hat{x}\\\hat{y}\end{bmatrix}=e_1x+e_2y  \rightarrow new basis
$$
$$\begin{bmatrix}0&1\\3&-1\end{bmatrix}\begin{bmatrix}\hat{x}\\\hat{y}\end{bmatrix}=\begin{bmatrix}7\\-16\end{bmatrix}$$
solution:
$$
\begin{bmatrix}\hat{x}\\\hat{y}\end{bmatrix}=\begin{bmatrix}1/3&1/3\\1&0\end{bmatrix}\begin{bmatrix}7\\-16\end{bmatrix}=\begin{bmatrix}-3\\7\end{bmatrix}
$$
Which means, minus three times on $e_1$ and seven times on $e_2$
A: I presume that you mean you have a new coordinate system, (x',y') where x'= x+ y and y'= -x+ y.  Adding the two equations, x'+ y'= 2y so y= (x'+ y')/2 and then x= x'- y= x'- (x'+ y')/2= x'/2- y'/2.
So x+ y= x'/2- y'/2+ x'/2+ y'/2= x'= 7 and
2x- 4y= 2(x'/2- y'/2)+ 4(x'/2+ y'/2)= x'- y'+ 2x'+ 2y'= 3x'- y'= -16.
The system of equations in this new basis is
x'= 7
3x'- y'= -16.
