Transformation matrices in $\mathbb{R}^2 \to \mathbb{R}^2$ Let $v_1=(1,1)$ and $v_2=(-1,1)$ vectors in $\mathbb{R}^2$. They are clearly linearly independent since each is not an scalar multiple of the other. The following information about a linear transformation $f: \mathbb{R}^2 \to \mathbb{R}^2$ is given: $$f(v_1)=10 \cdot v_1 \text{ and } f(v_2)=4 \cdot v_2$$


*

*Give the transformation matrix $_vF_v$ with respect to ordered basis $\mathcal{B}=(v_1,v_2)$

*Give the transformation matrix $_eF_e$ with respect to the ordered standard basis $e=(e_1,e_2)$ of $\mathbb{R}^2$
Recall that 
$$ \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} $$
We need a matrix $_eF_e$ such that:
$$_eF_e\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 10 & 0 \\ 0 & 4 \end{bmatrix}$$
then
$$_eF_e=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}^{-1}
=\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix}\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}=\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix}$$
Okay so I'm pretty sure that $$_eF_e=_eF_v \cdot _vF_v \cdot _vF_e$$
And i figured I could find $_eF_e=\begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix}$ in the following equation $$\begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix} \text{ } \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}= \begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix} \\ \Rightarrow
\begin{bmatrix} 10 & -4 \\ 10 & 4 \end{bmatrix} \text{ } \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}= \begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} \\ \Rightarrow {}_eF_e=\begin{bmatrix} 7 & 3 \\ 3 & 7 \end{bmatrix} $$
Now, how can I find ${}_v{F}_v$? I got a feeling that I'm making it more difficult than necessary

Recall that $v_1$ and $v_2$ are linearly independent eigenvectors of $f$ since:
$$f(v_1)=10\cdot v_1$$
$$f(v_2)=4\cdot v_2$$

The vectors $v_1$ and $v_2$ form a basis of $\mathbb{R}^2$ consisting of eigenvectors of $f$. So $_vF_v$ is similar to a diagonal matrix whose elements in the diagonal are the eigenvalues $\lambda_1=10$ and $\lambda_2=4$ of $f$.

So your matrix $_vF_v$ that represents $f$ w.r.t. the ordered basis $\mathcal{B}=(v_1,v_2)$ should be the diagonal form bellow:
$$_vF_v=\left[\begin{array}{cc} 10&0\\ 0&4\end{array}\right]$$
Also, if you let $P$ to be the matrix whose columns are $v_1$ and $v_2$:
$$P=\left[\begin{array}{c} \\ \ v_1\\ \\ \end{array}  \middle\vert \begin{array}{c} \\ \ v_2\\  \\ \end{array}\right]=\left[\begin{array}{cc} 1&-1\\ &\\1& \ 1\end{array}\right]$$
then
$$_{e}F_e = P \cdot{}_vF_v\cdot P^{-1}$$
or you can do this:
$$(x,y)=\alpha\cdot v_{1}+\beta\cdot v_{2}
=\alpha\cdot(1,1)+\beta\cdot(-1,1)=(\alpha-\beta,\alpha+\beta)$$
so
$$\alpha=\frac{x+y}{2}$$
$$\beta=\frac{y-x}{2}$$
then:
$$ (x,y)= \frac{(x+y)}{2}\cdot v_1 + \frac{(y-x)}{2}\cdot v_2$$
Apply $f$ in both sides to get:
$$f(x,y) = \frac{(x+y)}{2}\cdot f(v_1) + \frac{(y-x)}{2}\cdot f(v_2) $$
$$f(x,y) = \frac{(x+y)}{2}\cdot 10\cdot v_1 + \frac{(y-x)}{2}\cdot 4\cdot v_2$$
$$ = 5(x+y)\cdot v_1+2(y-x)\cdot v_2$$
$$ = (5x+5y,5x+5y)+(2x-2y,2y-2x)=(7x+3y,7y+3x)=(7x+3y,3x+7y)$$
So
$$f(x,y)=(7x+3y,3x+7y)$$
then
$$f(1,0)=(7,3)$$
$$f(0,1)=(3,7)$$
this implies that:
$$_{e}F_e = \left[\begin{array}{cc} 7&3\\ &\\3& 7\end{array}\right] $$
