Understanding a textbook example of linear mappings as matrices. I have a linear mapping $f:Q_{\leq1}[t] \to Q_{\leq1}[t]$ matrix A given by
$f(a_{1}t + a_{0}) = 2a_{1} + a_{0}$
And two bases given by
$B_{1} = \{1, t\}$ and $B_{2} = \{t+1, t-1\}$
so that
$[f]_{B1,B1} = \begin{bmatrix}
1 & 0\\
0 & 2
\end{bmatrix}$
$[f]_{B1,B2} = \begin{bmatrix}
1/2 & 1\\
-1/2 & 1
\end{bmatrix}$
$[f]_{B2,B2} = \begin{bmatrix}
3/2 & 1/2\\
1/2 & 3/2
\end{bmatrix}$
$[f]_{B2,B1} = \begin{bmatrix}
1 & -1\\
2 & 2
\end{bmatrix}$
I know this is really basic, but I am having trouble understanding how these mappings are constructed from B1 and B2. For the first one, I've figured out that it is clearly in relation to the identity matrix as a representation of B1, but beyond that I am a bit lost.
This topic is very important for me so please reply comprehensively.
 A: So, (I take it it's) $f(a_0+a_1t)=a_0+2a_1t$, or $f(\begin{pmatrix}a_0\\a_1\end{pmatrix})=\begin{pmatrix}a_0\\2a_1\end{pmatrix}$..
Once you have the first one, which you get by just applying $f$ to the elements of $B_1=\{\begin {pmatrix}1\\0\end {pmatrix},\begin {pmatrix}0\\1\end {pmatrix}\}$ and writing the results in terms of $B_1$, you get the others by applying the transition matrix (or its inverse).
The transition matrix between $B_1$ and $B_2$ is
$$P=[I]_{B_2,B_1}=\begin {pmatrix}1\quad -1\\1\quad 1\end {pmatrix}$$.  Notice that the columns are the elements of $B_2$.
Then $P^{-1}=\begin {pmatrix}1/2\quad 1/2\\-1/2\quad 1/2\end {pmatrix}$.
Thus $[f]_{B_1,B_2}=P^{-1}[f]_{B_1,B_1}$.
And $[f]_{B_2,B_1}=[f]_{B_1,B_1}P$.
Finally $[f]_{B_2,B_2}=P^{-1}[f]_{B_1,B_1}P$.
A: $f(a_1t+a_0)=2a_1t+a_0.$
To compute the first column of $[f]_{B_i,B_j}$, take the first vector of $B_i$, then its image by $f$, and finally the (two) coordinates of this image in $B_j.$ Same for the second column, starting from the second vector of $B_i.$
The coordinates of $a+bt$ in $B_1$ are $\begin{pmatrix}a\\b\end{pmatrix}.$
Since $1=\frac{(t+1)-(t-1)}2$ and $t=\frac{(t+1)+(t-1)}2,$ the coordinates in $B_2$ of $a+bt=a\frac{(t+1)-(t-1)}2+b\frac{(t+1)+(t-1)}2=\frac{a+b}2(t+1)+\frac{-a+b}2(t-1)$ are $\begin{pmatrix}\frac{a+b}2\\\frac{-a+b}2\end{pmatrix}.$
I think now you will be able to obtain your four matrices.
