How to find bases $\alpha$ and $\beta$ for the following $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$? Suppose a linear transformation $T:P_3(\Bbb{R})\rightarrow P_2(\Bbb{R})$ has the matrix $A=\begin{bmatrix}
    1 &2 & 0 & 0 \\
    0 & 1 & 2& 1 \\
    1 & 1 & 1 & 1 \\
\end{bmatrix} $ relative to the standard bases of $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$. How to to find bases $\alpha$ of $P_3(\Bbb{R})$ and $\beta$ for $P_2(\Bbb{R})$ such 
that the matrix $T$ relative to $\alpha$ and $\beta$ is the reduced row-echelon form of $A$?
I assume we have to use change of basis formula. I have difficulty to start this type of question 
because it's hard to identity the $[I]$ and $[T]$. Here the question says $A$ is relative to the standard 
bases of $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$ so I think $[\alpha]_\gamma$ and $[\beta]_\gamma$ are respectively $[1, x, x^2, x^3]$ and $[1, x, x^2]$. But 
then I don't know how to continue because I am not sure how $A$ works and how to write the change 
of basis formula.
 A: If you subtract to the third line the fist line and sum the second you get the matrix 
$$\begin{pmatrix} 1 & 2 & 0 & 0\\ 0 & 1 &2 &1 \\0&0&3&2 \end{pmatrix}=
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 &0  \\-1 &1&1 \end{pmatrix}
\begin{pmatrix} 1 & 2 & 0 & 0\\ 0 & 1 &2 &1 \\1&1&1&1 \end{pmatrix}$$ 
the matrix  $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 &0  \\-1 &1&1 \end{pmatrix}$ is the matrix of the identity application from the canonical basis to the basis you want (in the codomain, because t's on the left of $A$. You don't need to change the basis in the domain, because there's no need to work on the columns therefore not to multiply for a matrix on the right).
Then to find the new basis in terms of $1,x,x^2$ find the inverse (that is, the matrix of the identity application from the new basis to the canonical one) which is
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 &0  \\1 &-1&1 \end{pmatrix}$$
 As you can see from this matrix, the new basis is $\beta =\{1+x^2,x-x^2,x^2 \}$.
A: Note that for any matrix $A$ there is an invertible matrix $E$ such that 
$\DeclareMathOperator{rref}{rref}EA=\rref A$. One computes $E$ by keeping track of the row reductions. In our case, we have
\begin{align*}
A &=
\left[\begin{array}{rrrr}
1 & 2 & 0 & 0 \\
0 & 1 & 2 & 1 \\
1 & 1 & 1 & 1
\end{array}\right] &
E &=
\left[\begin{array}{rrr}
-\frac{1}{3} & -\frac{2}{3} & \frac{4}{3} \\
\frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\
-\frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{array}\right] &
\rref A &=
\left[\begin{array}{rrrr}
1 & 0 & 0 & \frac{2}{3} \\
0 & 1 & 0 & -\frac{1}{3} \\
0 & 0 & 1 & \frac{2}{3}
\end{array}\right]
\end{align*}
Note that $E$ is invertible, being the product of elementary matrices. This implies that $E$ is the change of basis matrix corresponding to the basis
\begin{align*}
p_1 &= -\frac{1}{3}+\frac{2}{3}\,x-\frac{1}{3}\,x^2 &
p_2 &= -\frac{2}{3}+\frac{1}{3}\,x+\frac{1}{3}\,x^2 &
p_3 &= \frac{4}{3}-\frac{2}{3}\,x+\frac{1}{3}\,x^2 
\end{align*}
