Change of basis for linear transformation - Linear algebra so i'm having a lot of difficulties with change of basis. Watched tons of tutorials on youtube but they only seem to confuse me more. 

Let $T: \mathbb{R^2} \to \mathbb{R^2}$ be defined by $T(a,b) = (a +
 2b, 3a - b)$. Let $\mathcal{B} = \{(1,1),(1,0)\}$ and $\mathcal{C} =
> \{(4,7),(4,8)\}$. Find $[T]_\mathcal{B}$ and $[T]_\mathcal{C}$ and
  show that $[T]_\mathcal{C} = Q^{-1} \cdot [T]_\mathcal{B}\cdot Q$ for
  some invertible matrix $Q$.

So i've done some thinking, and the matrix $Q$ might be the matrix that goes from basis $c$ to $b$? and then if i invert that one i get matrix $Q^{-1}$? I just have 0 idea where to start to problem from, like what's the first matrix that i have to work with?
Do i start with the basic basis in $R^2$ that is $\{(1,0),(0,1)\}$ and apply the transformation to it? 
 A: Take a look here
How do I express ordered bases for polynomials as a matrices? Linear Algebra.
In your case you know the matrix for the canonical basis:
$$T_A=\begin{bmatrix}
    1 & 2 \\
    3 & -1 
    \end{bmatrix}$$
And two other basis:
$$\mathcal{B} = \{(1,1),(1,0)\}$$ and 
$$\mathcal{C} = \{(4,7),(4,8)\}$$
and have to find the matrices $T_{B}$ and $T_{C}$ with respect these two basis.
You can follow the previous example to do it.

Let's see for example how to find out $T_B$.

Let be $w$ any vector given in the canonical basis and indicate with: $v_1=(1,1)$ and $v_2=(1,0)$ the two vectors of the basis $\mathcal{B}$, also given with respect to the canonical basis.
We are loking for the coefficient $x_1$ and $x_2$ such that:
$$w=x_1\cdot v_1+x_2\cdot v_2$$
or in matrix form:
$$w=V\cdot x$$
$$V=\begin{bmatrix}
    1 & 1 \\
    1 & 0 
    \end{bmatrix}$$
Note that $V$ contains as column the vectors of the new basis with respect to the canonical basis. It is important to note that $V$ represent the matrix of change of basis from $\mathcal{B}$ to the canonical.
Thus, the components of any vector $w$ with respect to the new basis are given by:
$$x=V^{-1}\cdot w$$
$$V^{-1}=\begin{bmatrix}
    0 & 1 \\
    1 & -1 
    \end{bmatrix}$$
Thus $V^{-1}$ represent the matrix of change of basis from the canonical to the new basis $\mathcal{B}$ .
The mapping of $w$ with respect to the canonical basis is given by:
$$u=T_A\cdot w$$
Since we are looking for the matrix $T_B$ that represent $T$ with respect to the new basis  $\mathcal{B}$, let's plug in the change of basis:
$u=V \cdot y$
$w=V \cdot x$
to obtain:
$$u=T_A\cdot w \iff V \cdot y=T_A\cdot V \cdot x \iff y=V^{-1}T_A\cdot V \cdot x=T_B\cdot x$$

Thus the matrix :
$$T_B=V^{-1}\cdot T_A\cdot V$$
represent the transformation with respect to the new basis $\mathcal{B}$.

For $T_C$ you can proceed in the same manner finding: $$T_C=W^{-1}\cdot T_A\cdot W$$ Now since $$T_B=V^{-1} \cdot T_A\cdot V \implies T_A=V \cdot T_B\cdot V^{-1}$$ you conclude $$T_C=W^{-1} \cdot T_A\cdot W=W^{-1} \cdot V \cdot T_B\cdot V^{-1}\cdot W$$
