I not really sure if I'm doing it right so I wanted to ask here:
I'm given a linear map $f:\Bbb{R^3} \rightarrow \Bbb{R^2} $ with $f((x,y,z)^T)= (x+y,y+z)^T$
and Bases $B=((1,-1,0)^T,(0,1,-1)^T,(1,1,1)^T)$ and $B'=((1,1)^T,(1,0)^T)$. I have to find the matrix of $f$ regarding the basis $B$ and $B'$
So I first wanted to know what f does to the standard basis of $\Bbb{R^3}$ and I got the Matrix
$$A=\begin{pmatrix}
1 & 1 & 0 \\\
0 & 1 & 1
\end{pmatrix}
$$
So next thing is to find both base transformation matrices.
If I want to know how i get from the standard basis to the basis $B$ I simply have to right down the Vectors in a Matrix:
$$M=\begin{pmatrix}
1 & 0 & 1 \\\
-1 & 1 & 1 \\\
0 & -1 & 1
\end{pmatrix}
$$
and the same goes for $B'$:
$$T=\begin{pmatrix}
1 & 1 \\\
1 & 0
\end{pmatrix}
$$
And then Multiply:
$$TAM=\begin{pmatrix}
1 & 1 \\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 1 & 0 \\\
0 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 1 \\\
-1 & 1 & 1 \\\
0 & -1 & 1
\end{pmatrix}
$$
$$=\begin{pmatrix}
-1 & 1 & 4 \\\
0 & 1 & 2
\end{pmatrix}
$$
Is that about right? I'm really unsure about it.
Best Answer
Not exactly, but almost.
You have to take the inverse of $T$.
Nevertheless, you can verify your solution by calculating $Ab_i$ (which are just the columns of $AM$) and expressing them in basis $B'$.
(This last step is equivalent to multiplying by $T^{-1}$ from the left. Can you see why?)